Introduction to Segment Trees – Data Structure and Algorithm Tutorials
Last Updated : 14 Dec, 2024
A Segment Tree is used to stores information about array intervals in its nodes.
- It allows efficient range queries over array intervals.
- Along with queries, it allows efficient updates of array items.
- For example, we can perform a range summation of an array between the range L to R in O(Log n) while also modifying any array element in O(Log n) Time
Types of Operations:
Segment Trees are mainly used for range queries on a fixed sized array. The values of array elements can be changed. The type of range queries that a segment tree can perform must have the property that to answer query from L to R, we can use the answers from L to M and M + 1 to R where M is a point between L to R, i.e., L <= M <= R.
The following are few example queries over a range in an array.
- Addition/Subtraction
- Maximum/Minimum
- GCD/LCM
- AND/OR/XOR
Structure of the Tree
The segment tree works on the principle of divide and conquer.
- At each level, we divide the array segments into two parts. If the given array had [0, . . ., N-1] elements in it then the two parts of the array will be [0, . . ., N/2-1] and [N/2, . . ., N-1].
- We will then recursively go on until the lower and upper bounds of the range become equal.
- The structure of the segment tree looks like a binary tree.
The segment tree is generally represented using an array where the first value stores the value for the total array range and the child of the node at the ith index are at (2*i + 1) and (2*i + 2).
Constructing the segment tree:
There are two important points to be noted while constructing the segment tree:
- Choosing what value to be stored in the nodes according to the problem definition
- What should the merge operation do
If the problem definition states that we need to calculate the sum over ranges, then the value at nodes should store the sum of values over the ranges.
- The child node values are merged back into the parent node to hold the value for that particular range, [i.e., the range covered by all the nodes of its subtree].
- In the end, leaf nodes store information about a single element. All the leaf nodes store the array based on which the segment tree is built.
Following are the steps for constructing a segment tree:
- Start from the leaves of the tree
- Recursively build the parents from the merge operation
The merge operation will take constant time if the operator takes constant time. SO building the whole tree takes O(N) time.
Segment tree
Range Query
Let us understand this with the help of the following problem
Given two integers L and R return the sum of the segment [L, R]
The first step is constructing the segment tree with the addition operator and 0 as the neutral element.
- If the range is one of the node’s range values then simply return the answer.
- Otherwise, we will need to traverse the left and right children of the nodes and recursively continue the process till we find a node that covers a range that totally covers a part or whole of the range [L, R]
- While returning from each call, we need to merge the answers received from each of its child.
As the height of the segment tree is logN the query time will be O(logN) per query.
Range Query in Segment Tree
Point Updates
Given an index, idx, update the value of the array at index idx with value V
The element’s contribution is only in the path from its leaf to its parent. Thus only logN elements will get affected due to the update.
For updating, traverse till the leaf that stores the value of index idx and update the value. Then while tracing back in the path, modify the ranges accordingly.
The time complexity will be O(logN).
Point Update in Segment Tree
Below is the implementation of construction, query, and point update for a segment tree:
C++ // C++ code for segment tree with sum // range and update query #include <bits/stdc++.h> using namespace std; vector<int> A, ST; void build(int node, int L, int R) { // Leaf node where L == R if (L == R) { ST[node] = A[L]; } else { // Find the middle element to // split the array into two halves int mid = (L + R) / 2; // Recursively travel the // left half build(2 * node, L, mid); // Recursively travel the // right half build(2 * node + 1, mid + 1, R); // Storing the sum of both the // children into the parent ST[node] = ST[2 * node] + ST[2 * node + 1]; } } void update(int node, int L, int R, int idx, int val) { // Find the lead node and // update its value if (L == R) { A[idx] += val; ST[node] += val; } else { // Find the mid int mid = (L + R) / 2; // If node value idx is at the // left part then update // the left part if (L <= idx and idx <= mid) update(2 * node, L, mid, idx, val); else update(2 * node + 1, mid + 1, R, idx, val); // Store the information in parents ST[node] = ST[2 * node] + ST[2 * node + 1]; } } int query(int node, int tl, int tr, int l, int r) { // If it lies out of range then // return 0 if (r < tl or tr < l) return 0; // If the node contains the range then // return the node value if (l <= tl and tr <= r) return ST[node]; int tm = (tl + tr) / 2; // Recursively traverse left and right // and find the node return query(2 * node, tl, tm, l, r) + query(2 * node + 1, tm + 1, tr, l, r); } // Driver code int main() { int n = 6; A = { 0, 1, 3, 5, -2, 3 }; // Create a segment tree of size 4*n ST.resize(4 * n); // Build a segment tree build(1, 0, n - 1); cout << "Sum of values in range 0-4 are: " << query(1, 0, n - 1, 0, 4) << "\n"; // Update the value at idx = 1 by // 100 thus becoming 101 update(1, 0, n - 1, 1, 100); cout << "Value at index 1 increased by 100\n"; cout << "sum of value in range 1-3 are: " << query(1, 0, n - 1, 1, 3) << "\n"; return 0; } // Java code for segment tree with sum // range and update query import java.io.*; import java.util.*; class GFG { static int n = 6; static int A[] = { 0, 1, 3, 5, -2, 3 }; // Create a segment tree of size 4*n static int ST[] = new int[4 * n]; public static void build(int node, int L, int R) { // Leaf node where L == R if (L == R) { ST[node] = A[L]; } else { // Find the middle element to // split the array into two halves int mid = (L + R) / 2; // Recursively travel the // left half build(2 * node, L, mid); // Recursively travel the // right half build(2 * node + 1, mid + 1, R); // Storing the sum of both the // children into the parent ST[node] = ST[2 * node] + ST[2 * node + 1]; } } public static void update(int node, int L, int R, int idx, int val) { // Find the lead node and // update its value if (L == R) { A[idx] += val; ST[node] += val; } else { // Find the mid int mid = (L + R) / 2; // If node value idx is at the // left part then update // the left part if (L <= idx && idx <= mid) update(2 * node, L, mid, idx, val); else update(2 * node + 1, mid + 1, R, idx, val); // Store the information in parents ST[node] = ST[2 * node] + ST[2 * node + 1]; } } public static int query(int node, int tl, int tr, int l, int r) { // If it lies out of range then // return 0 if (r < tl || tr < l) return 0; // If the node contains the range then // return the node value if (l <= tl && tr <= r) return ST[node]; int tm = (tl + tr) / 2; // Recursively traverse left and right // and find the node return query(2 * node, tl, tm, l, r) + query(2 * node + 1, tm + 1, tr, l, r); } // Driver Code public static void main(String[] args) { // Build a segment tree build(1, 0, n - 1); System.out.println( "Sum of values in range 0-4 are: " + query(1, 0, n - 1, 0, 4)); // Update the value at idx = 1 by // 100 ths becoming 101 update(1, 0, n - 1, 1, 100); System.out.println( "Value at index 1 increased by 100"); System.out.println("sum of value in range 1-3 are: " + query(1, 0, n - 1, 1, 3)); } } // This code is contributed by Rohit Pradhan # python3 code for segment tree with sum # range and update query A = [] ST = [] def build(node, L, R): global A, ST # Leaf node where L == R if (L == R): ST[node] = A[L] else: # Find the middle element to # split the array into two halves mid = (L + R) // 2 # Recursively travel the # left half build(2 * node, L, mid) # Recursively travel the # right half build(2 * node + 1, mid + 1, R) # Storing the sum of both the # children into the parent ST[node] = ST[2 * node] + ST[2 * node + 1] def update(node, L, R, idx, val): global A, ST # Find the lead node and # update its value if (L == R): A[idx] += val ST[node] += val else: # Find the mid mid = (L + R) // 2 # If node value idx is at the # left part then update # the left part if (L <= idx and idx <= mid): update(2 * node, L, mid, idx, val) else: update(2 * node + 1, mid + 1, R, idx, val) # Store the information in parents ST[node] = ST[2 * node] + ST[2 * node + 1] def query(node, tl, tr, l, r): global A, ST # If it lies out of range then # return 0 if (r < tl or tr < l): return 0 # If the node contains the range then # return the node value if (l <= tl and tr <= r): return ST[node] tm = (tl + tr) // 2 # Recursively traverse left and right # and find the node return query(2 * node, tl, tm, l, r) + query(2 * node + 1, tm + 1, tr, l, r) # Driver code if __name__ == "__main__": n = 6 A = [0, 1, 3, 5, -2, 3] # Create a segment tree of size 4*n ST = [0 for _ in range(4 * n)] # Build a segment tree build(1, 0, n - 1) print(f"Sum of values in range 0-4 are: {query(1, 0, n - 1, 0, 4)}") # Update the value at idx = 1 by # 100 ths becoming 101 update(1, 0, n - 1, 1, 100) print("Value at index 1 increased by 100") print(f"sum of value in range 1-3 are: {query(1, 0, n - 1, 1, 3)}") # This code is contributed by rakeshsahni // C# code for segment tree with sum // range and update query using System; public class GFG { static int n = 6; static int[] A = { 0, 1, 3, 5, -2, 3 }; // Create a segment tree of size 4*n static int[] ST = new int[4 * n]; public static void build(int node, int L, int R) { // Leaf node where L == R if (L == R) { ST[node] = A[L]; } else { // Find the middle element to // split the array into two halves int mid = (L + R) / 2; // Recursively travel the // left half build(2 * node, L, mid); // Recursively travel the // right half build(2 * node + 1, mid + 1, R); // Storing the sum of both the // children into the parent ST[node] = ST[2 * node] + ST[2 * node + 1]; } } public static void update(int node, int L, int R, int idx, int val) { // Find the lead node and // update its value if (L == R) { A[idx] += val; ST[node] += val; } else { // Find the mid int mid = (L + R) / 2; // If node value idx is at the // left part then update // the left part if (L <= idx && idx <= mid) update(2 * node, L, mid, idx, val); else update(2 * node + 1, mid + 1, R, idx, val); // Store the information in parents ST[node] = ST[2 * node] + ST[2 * node + 1]; } } public static int query(int node, int tl, int tr, int l, int r) { // If it lies out of range then // return 0 if (r < tl || tr < l) return 0; // If the node contains the range then // return the node value if (l <= tl && tr <= r) return ST[node]; int tm = (tl + tr) / 2; // Recursively traverse left and right // and find the node return query(2 * node, tl, tm, l, r) + query(2 * node + 1, tm + 1, tr, l, r); } static public void Main() { // Code // Build a segment tree build(1, 0, n - 1); Console.WriteLine("Sum of values in range 0-4 are: " + query(1, 0, n - 1, 0, 4)); // Update the value at idx = 1 by // 100 ths becoming 101 update(1, 0, n - 1, 1, 100); Console.WriteLine( "Value at index 1 increased by 100"); Console.WriteLine("sum of value in range 1-3 are: " + query(1, 0, n - 1, 1, 3)); } } // This code is contributed by lokeshmvs21. // JavaScript code for the above approach function build(node, L, R) { // Leaf node where L == R if (L === R) { ST[node] = A[L]; } else { // Find the middle element to split the array into two halves const mid = Math.floor((L + R) / 2); // Recursively travel the left half build(2 * node, L, mid); // Recursively travel the right half build(2 * node + 1, mid + 1, R); // Storing the sum of both the children into the parent ST[node] = ST[2 * node] + ST[2 * node + 1]; } } function update(node, L, R, idx, val) { // Find the lead node and update its value if (L === R) { A[idx] += val; ST[node] += val; } else { // Find the mid const mid = Math.floor((L + R) / 2); // If node value idx is at the left part then update the left part if (L <= idx && idx <= mid) { update(2 * node, L, mid, idx, val); } else { update(2 * node + 1, mid + 1, R, idx, val); } // Store the information in parents ST[node] = ST[2 * node] + ST[2 * node + 1]; } } function query(node, tl, tr, l, r) { // If it lies out of range then return 0 if (r < tl || tr < l) return 0; // If the node contains the range then return the node value if (l <= tl && tr <= r) return ST[node]; const tm = Math.floor((tl + tr) / 2); // Recursively traverse left and right and find the node return query(2 * node, tl, tm, l, r) + query(2 * node + 1, tm + 1, tr, l, r); } // Driver Code const A = [0, 1, 3, 5, -2, 3]; const ST = new Array(4 * A.length); // Build a segment tree build(1, 0, A.length - 1); console.log(`Sum of values in range 0-4 are: ${query(1, 0, A.length - 1, 0, 4)}<br>`); // Update the value at idx = 1 by 100 thus becoming 101 update(1, 0, A.length - 1, 1, 100); console.log(`Value at index 1 increased by 100 <br>`); console.log(`sum of value in range 1-3 are: ${query(1, 0, A.length - 1, 1, 3)}<br>`); // This code is contributed by Potta Lokesh OutputSum of values in range 0-4 are: 7 Value at index 1 increased by 100 sum of value in range 1-3 are: 109
Time complexity: O(N)
- The building operation takes O(N) time
- The query operation takes O(logN) time
- Each update is performed in O(logN) time
Auxiliary Space: O(n)
Note:
A segment tree with 2^x leaf nodes will have 2^(x+1)-1 total nodes due to its perfect binary tree structure. However, when dealing with a non-power-of-two number of elements, extra leaf nodes may be present. To represent all elements, the number of leaf nodes must be rounded up to the nearest power of two, resulting in a maximum of almost 2*n leaf nodes.
For instance, if n is 2^j + 1, 2^(j+1) leaf nodes will be required, leading to an O(2*n) space complexity. As the total number of nodes is about twice the number of leaf nodes, the total space complexity of the segment tree is O(4n). The space requirement can be substantial, but it is usually manageable for most practical applications.
Updating an interval (Lazy propagation):
Lazy Propagation: A speedup technique for range updates
Example of segment tree
- We can delay some updates (avoid recursive calls in update) and do such updates only when necessary when there are several updates and updates are being performed on a range.
- A node in a segment tree stores or displays the results of a query for a variety of indexes.
- Additionally, all of the node’s descendants must also be updated if the update operation’s range includes this node.
- Take the node with the value 27 in the picture above as an example. This node contains the sum of values at the indexes 3 to 5. This node and all of its descendants must be updated if our update query covers the range of 2 to 5.
- By storing this update information in distinct nodes referred to as lazy nodes or values, we use lazy propagation to update only the node with value 27 and delay updates to its descendants.
- We make an array called lazy[] to stand in for the lazy node. The size of lazy[] is the same as the array used to represent the segment tree in the code following, which is tree[].
- The goal is to set all of the lazy[elements] to 0.
- There are no pending changes on the segment tree node i if lazy[i] has a value of 0.
- A non-zero value for lazy[i] indicates that before doing any queries on node i in the segment tree, this sum needs to be added to the node.
Below is the implementation to demonstrate the working of Lazy Propagation.
C++ // Program to show segment tree to // demonstrate lazy propagation #include <bits/stdc++.h> using namespace std; #define MAX 1000 // Ideally, we should not use global // variables and large constant-sized // arrays, we have done it here for // simplicity. // To store segment tree int tree[MAX] = { 0 }; // To store pending updates int lazy[MAX] = { 0 }; // si -> index of current node in segment tree // ss and se -> Starting and ending // indices of elements for which current // nodes stores sum. // us and ue -> starting and ending indexes // of update query // diff -> which we need to add in the // range us to ue void updateRangeUtil(int si, int ss, int se, int us, int ue, int diff) { // If lazy value is non-zero for // current node of segment tree, then // there are some pending updates. So, // we need to make sure that the // pending updates are done before // making new updates. Because this // value may be used by parent after // recursive calls (See last line // of this function) if (lazy[si] != 0) { // Make pending updates using // value stored in lazy nodes tree[si] += (se - ss + 1) * lazy[si]; // checking if it is not leaf node // because if it is leaf node then // we cannot go further if (ss != se) { // We can postpone updating // children we don't need // their new values now. Since // we are not yet updating // children of si, we need to // set lazy flags for the children lazy[si * 2 + 1] += lazy[si]; lazy[si * 2 + 2] += lazy[si]; } // Set the lazy value for current // node as 0 as it has been updated lazy[si] = 0; } // out of range if (ss > se || ss > ue || se < us) return; // Current segment is fully in range if (ss >= us && se <= ue) { // Add the difference to // current node tree[si] += (se - ss + 1) * diff; // Same logic for checking // leaf node or not if (ss != se) { // This is where we store // values in lazy nodes, // rather than updating the // segment tree itself Since // we don't need these updated // values now we postpone // updates by storing values // in lazy[] lazy[si * 2 + 1] += diff; lazy[si * 2 + 2] += diff; } return; } // If not completely in rang, // but overlaps, recur for children, int mid = (ss + se) / 2; updateRangeUtil(si * 2 + 1, ss, mid, us, ue, diff); updateRangeUtil(si * 2 + 2, mid + 1, se, us, ue, diff); // And use the result of children // calls to update this node tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2]; } // Function to update a range of values // in segment tree // us and eu -> starting and ending // indices of update query // ue -> ending index of update query // diff -> which we need to add in the // range us to ue void updateRange(int n, int us, int ue, int diff) { updateRangeUtil(0, 0, n - 1, us, ue, diff); } // A recursive function to get the sum of // values in given range of the array. // The following are parameters for this function. // si --> Index of current node in the st. // Initially 0 is passed as root is always // at' index 0 // ss & se --> Starting and ending indices // of the segment represented by current // node, i.e., tree[si] // qs & qe --> Starting and ending // indices of query range int getSumUtil(int ss, int se, int qs, int qe, int si) { // If lazy flag is set for current // node of segment tree, then there // are some pending updates. So we // need to make sure that the pending // updates are done before // processing the sub sum query if (lazy[si] != 0) { // Make pending updates to this // node. Note that this node // represents sum of elements in // arr[ss..se] and all these // elements must be increased by // lazy[si] tree[si] += (se - ss + 1) * lazy[si]; // Checking if it is not leaf node // because if it is leaf node then // we cannot go further if (ss != se) { // Since we are not yet // updating children os si, // we need to set lazy values // for the children lazy[si * 2 + 1] += lazy[si]; lazy[si * 2 + 2] += lazy[si]; } // unset the lazy value for current // node as it has been updated lazy[si] = 0; } // Out of range if (ss > se || ss > qe || se < qs) return 0; // At this point we are sure that // pending lazy updates are done for // current node. So we can return // value // If this segment lies in range if (ss >= qs && se <= qe) return tree[si]; // If a part of this segment overlaps // with the given range int mid = (ss + se) / 2; return getSumUtil(ss, mid, qs, qe, 2 * si + 1) + getSumUtil(mid + 1, se, qs, qe, 2 * si + 2); } // Return sum of elements in range from // index qs (query start) to qe (query end). // It mainly uses getSumUtil() int getSum(int n, int qs, int qe) { // Check for erroneous input values if (qs < 0 || qe > n - 1 || qs > qe) { cout << "Invalid Input"; return -1; } return getSumUtil(0, n - 1, qs, qe, 0); } // A recursive function that constructs // Segment Tree for array[ss..se]. // si is index of current node in st. void constructSTUtil(int arr[], int ss, int se, int si) { // Out of range as ss can never // be greater than se if (ss > se) return; // If there is one element in array, // store it in current node of segment // tree and return if (ss == se) { tree[si] = arr[ss]; return; } // If there are more than one elements, // then recur for left and right // subtrees and store the sum of // values in this node int mid = (ss + se) / 2; constructSTUtil(arr, ss, mid, si * 2 + 1); constructSTUtil(arr, mid + 1, se, si * 2 + 2); tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2]; } // Function to construct segment tree // from given array.This function allocates // memory for segment tree and calls // constructSTUtil() to fill the // allocated memory void constructST(int arr[], int n) { // Fill the allocated memory st constructSTUtil(arr, 0, n - 1, 0); } // Driver program to test above functions int main() { int arr[] = { 1, 3, 5, 7, 9, 11 }; int n = sizeof(arr) / sizeof(arr[0]); // Build segment tree from given array constructST(arr, n); // Print sum of values in array // from index 1 to 3 cout << "Sum of values in given range = " << getSum(n, 1, 3) << endl; // Add 10 to all nodes at indexes // from 1 to 5. updateRange(n, 1, 5, 10); // Find sum after the value is updated cout << "Updated sum of values in given range = " << getSum(n, 1, 3) << endl; return 0; } // Java program to show segment tree to // demonstrate lazy propagation import java.util.*; class GFG { // Ideally, we should not use global // variables and large constant-sized // arrays, we have done it here for // simplicity. static final int MAX = 1000; // To store segment tree static int[] tree = new int[MAX]; // To store pending updates static int[] lazy = new int[MAX]; // si -> index of current node in segment tree // ss and se -> Starting and ending // indices of elements for which current // nodes stores sum. // us and ue -> starting and ending indexes // of update query // diff -> which we need to add in the // range us to ue static void updateRangeUtil(int si, int ss, int se, int us, int ue, int diff) { // If lazy value is non-zero for // current node of segment tree, then // there are some pending updates. So, // we need to make sure that the // pending updates are done before // making new updates. Because this // value may be used by parent after // recursive calls (See last line // of this function) if (lazy[si] != 0) { // Make pending updates using // value stored in lazy nodes tree[si] += (se - ss + 1) * lazy[si]; // checking if it is not leaf node // because if it is leaf node then // we cannot go further if (ss != se) { // We can postpone updating // children we don't need // their new values now. Since // we are not yet updating // children of si, we need to // set lazy flags for the children lazy[si * 2 + 1] += lazy[si]; lazy[si * 2 + 2] += lazy[si]; } // Set the lazy value for current // node as 0 as it has been updated lazy[si] = 0; } // out of range if (ss > se || ss > ue || se < us) { return; } // Current segment is fully in range if (ss >= us && se <= ue) { // Add the difference to // current node tree[si] += (se - ss + 1) * diff; // Same logic for checking // leaf node or not if (ss != se) { // This is where we store // values in lazy nodes, // rather than updating the // segment tree itself Since // we don't need these updated // values now we postpone // updates by storing values // in lazy[] lazy[si * 2 + 1] += diff; lazy[si * 2 + 2] += diff; } return; } // If not completely in rang, // but overlaps, recur for children, int mid = (ss + se) / 2; updateRangeUtil(si * 2 + 1, ss, mid, us, ue, diff); updateRangeUtil(si * 2 + 2, mid + 1, se, us, ue, diff); // And use the result of children // calls to update this node tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2]; } // Function to update a range of values // in segment tree // us and eu -> starting and ending // indices of update query // ue -> ending index of update query // diff -> which we need to add in the // range us to ue static void updateRange(int n, int us, int ue, int diff) { updateRangeUtil(0, 0, n - 1, us, ue, diff); } // A recursive function to get the sum of // values in given range of the array. // The following are parameters for this function. // si --> Index of current node in the st. // Initially 0 is passed as root is always // at' index 0 // ss & se --> Starting and ending indices // of the segment represented by current // node, i.e., tree[si] // qs & qe --> Starting and ending // indices of query range static int getSumUtil(int ss, int se, int qs, int qe, int si) { // If lazy flag is set for current // node of segment tree, then there // are some pending updates. So we // need to make sure that the pending // updates are done before // processing the sub sum query if (lazy[si] != 0) { // Make pending updates to this // node. Note that this node // represents sum of elements in // arr[ss..se] and all these // elements must be increased by // lazy[si] tree[si] += (se - ss + 1) * lazy[si]; // Checking if it is not leaf node // because if it is leaf node then // we cannot go further if (ss != se) { // Since we are not yet // updating children os si, // we need to set lazy values // for the children lazy[si * 2 + 1] += lazy[si]; lazy[si * 2 + 2] += lazy[si]; } // unset the lazy value for current // node as it has been updated lazy[si] = 0; } // Out of range if (ss > se || ss > qe || se < qs) { return 0; } // At this point we are sure that // pending lazy updates are done for // current node. So we can return // value // If this segment lies in range if (ss >= qs && se <= qe) { return tree[si]; } // If a part of this segment overlaps // with the given range int mid = (ss + se) / 2; return getSumUtil(ss, mid, qs, qe, 2 * si + 1) + getSumUtil(mid + 1, se, qs, qe, 2 * si + 2); } // Return sum of elements in range from // index qs (query start) to qe (query end). // It mainly uses getSumUtil() static int getSum(int n, int qs, int qe) { // Check for erroneous input values if (qs < 0 || qe > n - 1 || qs > qe) { System.out.println("Invalid Input"); return -1; } return getSumUtil(0, n - 1, qs, qe, 0); } // A recursive function that constructs // Segment Tree for array[ss..se]. // si is index of current node in st. static void constructSTUtil(int arr[], int ss, int se, int si) { // Out of range as ss can never // be greater than se if (ss > se) return; // If there is one element in array, // store it in current node of segment // tree and return if (ss == se) { tree[si] = arr[ss]; return; } // If there are more than one elements, // then recur for left and right // subtrees and store the sum of // values in this node int mid = (ss + se) / 2; constructSTUtil(arr, ss, mid, si * 2 + 1); constructSTUtil(arr, mid + 1, se, si * 2 + 2); tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2]; } // Function to construct segment tree // from given array.This function allocates // memory for segment tree and calls // constructSTUtil() to fill the // allocated memory static void constructST(int arr[], int n) { // Fill the allocated memory st constructSTUtil(arr, 0, n - 1, 0); } // Driver program to test above functions public static void main(String[] args) { int arr[] = { 1, 3, 5, 7, 9, 11 }; int n = arr.length; // Build segment tree from given array constructST(arr, n); // Print sum of values in array // from index 1 to 3 System.out.println("Sum of values in given range = " + getSum(n, 1, 3)); // Add 10 to all nodes at indexes // from 1 to 5. updateRange(n, 1, 5, 10); // Find sum after the value is updated System.out.println( "Updated sum of values in given range = " + getSum(n, 1, 3)); } } // This code is contributed by Prasad Kandekar(prasad264) # Program to show segment tree to # demonstrate lazy propagation MAX = 1000 # Ideally, we should not use global # variables and large constant-sized # arrays, we have done it here for # simplicity. # To store segment tree tree = [0] * MAX # To store pending updates lazy = [0] * MAX # si -> index of current node in segment tree # ss and se -> Starting and ending # indices of elements for which current # nodes stores sum. # us and ue -> starting and ending indexes # of update query # diff -> which we need to add in the # range us to ue def updateRangeUtil(si, ss, se, us, ue, diff): # If lazy value is non-zero for # current node of segment tree, then # there are some pending updates. So, # we need to make sure that the # pending updates are done before # making new updates. Because this # value may be used by parent after # recursive calls (See last line # of this function) if lazy[si] != 0: # Make pending updates using # value stored in lazy nodes tree[si] += (se - ss + 1) * lazy[si] # checking if it is not leaf node # because if it is leaf node then # we cannot go further if ss != se: # We can postpone updating # children we don't need # their new values now. Since # we are not yet updating # children of si, we need to # set lazy flags for the children lazy[si * 2 + 1] += lazy[si] lazy[si * 2 + 2] += lazy[si] # Set the lazy value for current # node as 0 as it has been updated lazy[si] = 0 # out of range if ss > se or ss > ue or se < us: return # Current segment is fully in range if ss >= us and se <= ue: # Add the difference to # current node tree[si] += (se - ss + 1) * diff # Same logic for checking # leaf node or not if ss != se: # This is where we store # values in lazy nodes, # rather than updating the # segment tree itself Since # we don't need these updated # values now we postpone # updates by storing values # in lazy[] lazy[si * 2 + 1] += diff lazy[si * 2 + 2] += diff return # If not completely in rang, # but overlaps, recur for children, mid = (ss + se) // 2 updateRangeUtil(si * 2 + 1, ss, mid, us, ue, diff) updateRangeUtil(si * 2 + 2, mid + 1, se, us, ue, diff) # And use the result of children # calls to update this node tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2] # Function to update a range of values # in segment tree # us and eu -> starting and ending # indices of update query # ue -> ending index of update query # diff -> which we need to add in the # range us to ue def updateRange(n, us, ue, diff): updateRangeUtil(0, 0, n - 1, us, ue, diff) # A recursive function to get the sum of # values in given range of the array. # The following are parameters for this function. # si --> Index of current node in the st. # Initially 0 is passed as root is always # at' index 0 # ss & se --> Starting and ending indices # of the segment represented by current # node, i.e., tree[si] # qs & qe --> Starting and ending # indices of query range def getSumUtil(ss, se, qs, qe, si): # If lazy flag is set for current # node of segment tree, then there # are some pending updates. So we # need to make sure that the pending # updates are done before # processing the sub sum query if lazy[si] != 0: # Make pending updates to this # node. Note that this node # represents sum of elements in # arr[ss..se] and all these # elements must be increased by # lazy[si] tree[si] += (se - ss + 1) * lazy[si] # Checking if it is not leaf node # because if it is leaf node then # we cannot go further if ss != se: # Since we are not yet # updating children os si, # we need to set lazy values # for the children lazy[si * 2 + 1] += lazy[si] lazy[si * 2 + 2] += lazy[si] # unset the lazy value for current # node as it has been updated lazy[si] = 0 # Out of range if ss > se or ss > qe or se < qs: return 0 # At this point we are sure that # pending lazy updates are done for # current node. So we can return # value # If this segment lies in range if ss >= qs and se <= qe: return tree[si] # If a part of this segment overlaps # with the given range mid = (ss + se) // 2 return (getSumUtil(ss, mid, qs, qe, 2 * si + 1) + getSumUtil(mid + 1, se, qs, qe, 2 * si + 2)) # Return sum of elements in range from # index qs (query start) to qe (query end). # It mainly uses getSumUtil() def getSum(n, qs, qe): # Check for erroneous input values if qs < 0 or qe > n - 1 or qs > qe: print("Invalid Input") return -1 return getSumUtil(0, n - 1, qs, qe, 0) # A recursive function that constructs # Segment Tree for array[ss..se]. # si is index of current node in st. def constructSTUtil(arr, ss, se, si): # Out of range as ss can never # be greater than se if ss > se: return # If there is one element in array, # store it in current node of segment # tree and return if ss == se: tree[si] = arr[ss] return # If there are more than one elements, # then recur for left and right # subtrees and store the sum of # values in this node mid = (ss + se) // 2 constructSTUtil(arr, ss, mid, si * 2 + 1) constructSTUtil(arr, mid + 1, se, si * 2 + 2) tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2] # Function to construct segment tree # from given array.This function allocates # memory for segment tree and calls # constructSTUtil() to fill the # allocated memory def constructST(arr, n): # Fill the allocated memory st constructSTUtil(arr, 0, n - 1, 0) # Driver program to test above functions arr = [1, 3, 5, 7, 9, 11] n = len(arr) # Build segment tree from given array constructST(arr, n) # Print sum of values in array # from index 1 to 3 print(f"Sum of values in given range = {getSum(n, 1, 3)}") # Add 10 to all nodes at indexes # from 1 to 5. updateRange(n, 1, 5, 10) # Find sum after the value is updated print(f"Updated sum of values in given range = {getSum(n, 1, 3)}") // Program to show segment tree to // demonstrate lazy propagation using System; public class GFG { const int MAX = 1000; // Ideally, we should not use global // variables and large constant-sized // arrays, we have done it here for // simplicity. // To store segment tree static int[] tree = new int[MAX]; // To store pending updates static int[] lazy = new int[MAX]; // si -> index of current node in segment tree // ss and se -> Starting and ending // indices of elements for which current // nodes stores sum. // us and ue -> starting and ending indexes // of update query // diff -> which we need to add in the // range us to ue static void UpdateRangeUtil(int si, int ss, int se, int us, int ue, int diff) { // If lazy value is non-zero for // current node of segment tree, then // there are some pending updates. So, // we need to make sure that the // pending updates are done before // making new updates. Because this // value may be used by parent after // recursive calls (See last line // of this function) if (lazy[si] != 0) { // Make pending updates using // value stored in lazy nodes tree[si] += (se - ss + 1) * lazy[si]; // checking if it is not leaf node // because if it is leaf node then // we cannot go further if (ss != se) { // We can postpone updating // children we don't need // their new values now. Since // we are not yet updating // children of si, we need to // set lazy flags for the children lazy[si * 2 + 1] += lazy[si]; lazy[si * 2 + 2] += lazy[si]; } // Set the lazy value for current // node as 0 as it has been updated lazy[si] = 0; } // out of range if (ss > se || ss > ue || se < us) return; // Current segment is fully in range if (ss >= us && se <= ue) { // Add the difference to // current node tree[si] += (se - ss + 1) * diff; // Same logic for checking // leaf node or not if (ss != se) { // This is where we store // values in lazy nodes, // rather than updating the // segment tree itself Since // we don't need these updated // values now we postpone // updates by storing values // in lazy[] lazy[si * 2 + 1] += diff; lazy[si * 2 + 2] += diff; } return; } // If not completely in rang, // but overlaps, recur for children, int mid = (ss + se) / 2; UpdateRangeUtil(si * 2 + 1, ss, mid, us, ue, diff); UpdateRangeUtil(si * 2 + 2, mid + 1, se, us, ue, diff); // And use the result of children // calls to update this node tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2]; } // Function to update a range of values // in segment tree // us and eu -> starting and ending // indices of update query // ue -> ending index of update query // diff -> which we need to add in the // range us to ue static void UpdateRange(int n, int us, int ue, int diff) { UpdateRangeUtil(0, 0, n - 1, us, ue, diff); } // A recursive function to get the sum of // values in given range of the array. // The following are parameters for this function. // si --> Index of current node in the st. // Initially 0 is passed as root is always // at' index 0 // ss & se --> Starting and ending indices // of the segment represented by current // node, i.e., tree[si] // qs & qe --> Starting and ending // indices of query range static int GetSumUtil(int ss, int se, int qs, int qe, int si) { // If lazy flag is set for current // node of segment tree, then there // are some pending updates. So we // need to make sure that the pending // updates are done before // processing the sub sum query if (lazy[si] != 0) { // Make pending updates to this // node. Note that this node // represents sum of elements in // arr[ss..se] and all these // elements must be increased by // lazy[si] tree[si] += (se - ss + 1) * lazy[si]; // Checking if it is not leaf node // because if it is leaf node then // we cannot go further if (ss != se) { // Since we are not yet // updating children os si, // we need to set lazy values // for the children lazy[si * 2 + 1] += lazy[si]; lazy[si * 2 + 2] += lazy[si]; } // unset the lazy value for current // node as it has been updated lazy[si] = 0; } // Out of range if (ss > se || ss > qe || se < qs) return 0; // At this point we are sure that // pending lazy updates are done for // current node. So we can return // value // If this segment lies in range if (ss >= qs && se <= qe) return tree[si]; // If a part of this segment overlaps // with the given range int mid = (ss + se) / 2; return GetSumUtil(ss, mid, qs, qe, 2 * si + 1) + GetSumUtil(mid + 1, se, qs, qe, 2 * si + 2); } // Return sum of elements in range from // index qs (query start) to qe (query end). // It mainly uses getSumUtil() static int GetSum(int n, int qs, int qe) { // Check for erroneous input values if (qs < 0 || qe > n - 1 || qs > qe) { Console.WriteLine("Invalid Input"); return -1; } return GetSumUtil(0, n - 1, qs, qe, 0); } // A recursive function that constructs // Segment Tree for array[ss..se]. // si is index of current node in st. static void ConstructSTUtil(int[] arr, int ss, int se, int si) { // Out of range as ss can never // be greater than se if (ss > se) return; // If there is one element in array, // store it in current node of segment // tree and return if (ss == se) { tree[si] = arr[ss]; return; } // If there are more than one elements, // then recur for left and right // subtrees and store the sum of // values in this node int mid = (ss + se) / 2; ConstructSTUtil(arr, ss, mid, si * 2 + 1); ConstructSTUtil(arr, mid + 1, se, si * 2 + 2); tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2]; } // Function to construct segment tree // from given array.This function allocates // memory for segment tree and calls // constructSTUtil() to fill the // allocated memory static void ConstructST(int[] arr, int n) { // Fill the allocated memory st ConstructSTUtil(arr, 0, n - 1, 0); } // Driver program to test above functions static public void Main(string[] args) { int[] arr = { 1, 3, 5, 7, 9, 11 }; int n = arr.Length; // Build segment tree from given array ConstructST(arr, n); // Print sum of values in array // from index 1 to 3 Console.WriteLine("Sum of values in given range = " + GetSum(n, 1, 3)); // Add 10 to all nodes at indexes // from 1 to 5. UpdateRange(n, 1, 5, 10); // Find sum after the value is updated Console.WriteLine( "Updated sum of values in given range = " + GetSum(n, 1, 3)); } } // This code is contributed by Prasad Kandekar(prasad264) //javascript Program to show segment tree to // demonstrate lazy propagation let MAX = 1000; // Ideally, we should not use global // variables and large constant-sized // arrays, we have done it here for // simplicity. // To store segment tree let tree = new Array(MAX).fill(0); // To store pending updates let lazy = new Array(MAX).fill(0); // si -> index of current node in segment tree // ss and se -> Starting and ending // indices of elements for which current // nodes stores sum. // us and ue -> starting and ending indexes // of update query // diff -> which we need to add in the // range us to ue function updateRangeUtil( si,ss, se, us, ue, diff) { // console.log(si,ss,se,us,ue,diff); // If lazy value is non-zero for // current node of segment tree, then // there are some pending updates. So, // we need to make sure that the // pending updates are done before // making new updates. Because this // value may be used by parent after // recursive calls (See last line // of this function) if (lazy[si] != 0) { // Make pending updates using // value stored in lazy nodes tree[si] += (se - ss + 1) * lazy[si]; // checking if it is not leaf node // because if it is leaf node then // we cannot go further if (ss != se) { // We can postpone updating // children we don't need // their new values now. Since // we are not yet updating // children of si, we need to // set lazy flags for the children lazy[si * 2 + 1] += lazy[si]; lazy[si * 2 + 2] += lazy[si]; } // Set the lazy value for current // node as 0 as it has been updated lazy[si] = 0; } // out of range if (ss > se || ss > ue || se < us) return ; // Current segment is fully in range if (ss >= us && se <= ue) { // Add the difference to // current node tree[si] += (se - ss + 1) * diff; // Same logic for checking // leaf node or not if (ss != se) { // This is where we store // values in lazy nodes, // rather than updating the // segment tree itself Since // we don't need these updated // values now we postpone // updates by storing values // in lazy[] lazy[si * 2 + 1] += diff; lazy[si * 2 + 2] += diff; } return; } // If not completely in rang, // but overlaps, recur for children, let mid = (ss + se) / 2; updateRangeUtil(Math.floor(si * 2 + 1), Math.floor(ss), Math.floor(mid), Math.floor(us), Math.floor(ue), Math.floor(diff)); updateRangeUtil(Math.floor(si * 2 + 2), Math.floor(mid + 1), Math.floor(se), Math.floor(us), Math.floor(ue), Math.floor(diff)); // And use the result of children // calls to update this node tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2]; } // Function to update a range of values // in segment tree // us and eu -> starting and ending // indices of update query // ue -> ending index of update query // diff -> which we need to add in the // range us to ue function updateRange(n, us, ue, diff) { updateRangeUtil(0, 0, n - 1, us, ue, diff); } // A recursive function to get the sum of // values in given range of the array. // The following are parameters for this function. // si --> Index of current node in the st. // Initially 0 is passed as root is always // at' index 0 // ss & se --> Starting and ending indices // of the segment represented by current // node, i.e., tree[si] // qs & qe --> Starting and ending // indices of query range function getSumUtil( ss, se, qs, qe, si) { // console.log(ss,se,qs,qe,si); // // If lazy flag is set for current // node of segment tree, then there // are some pending updates. So we // need to make sure that the pending // updates are done before // processing the sub sum query if (lazy[si] != 0) { // Make pending updates to this // node. Note that this node // represents sum of elements in // arr[ss..se] and all these // elements must be increased by // lazy[si] tree[si] += (se - ss + 1) * lazy[si]; // Checking if it is not leaf node // because if it is leaf node then // we cannot go further if (ss != se) { // Since we are not yet // updating children os si, // we need to set lazy values // for the children lazy[si * 2 + 1] += lazy[si]; lazy[si * 2 + 2] += lazy[si]; } // unset the lazy value for current // node as it has been updated lazy[si] = 0; } // Out of range if (ss > se || ss > qe || se < qs) return 0; // At this point we are sure that // pending lazy updates are done for // current node. So we can return // value // If this segment lies in range if (ss >= qs && se <= qe) return tree[si]; // If a part of this segment overlaps // with the given range let mid = (ss + se) / 2; return getSumUtil(Math.floor(ss), Math.floor(mid), Math.floor(qs), Math.floor(qe), Math.floor(2 * si + 1)) + getSumUtil(Math.floor(mid + 1), Math.floor(se), Math.floor(qs), Math.floor(qe), Math.floor(2 * si + 2)); } // Return sum of elements in range from // index qs (query start) to qe (query end). // It mainly uses getSumUtil() function getSum (n, qs, qe) { // Check for erroneous input values if (qs < 0 || qe > n - 1 || qs > qe) { console.log("Invalid Input"); return -1; } // console.log(n,qs,qe); return getSumUtil(0, n - 1, qs, qe, 0); } // A recursive function that constructs // Segment Tree for array[ss..se]. // si is index of current node in st. function constructSTUtil( arr, ss, se, si) { // console.log(arr,ss,se,si); // Out of range as ss can never // be greater than se if (ss > se) return; // If there is one element in array, // store it in current node of segment // tree and return if (ss == se) { tree[si] = arr[ss]; return; } // If there are more than one elements, // then recur for left and right // subtrees and store the sum of // values in this node let mid = (ss + se) / 2; constructSTUtil(arr, Math.floor(ss), Math.floor(mid), Math.floor(si * 2 + 1)); constructSTUtil(arr, Math.floor(mid + 1), Math.floor(se), Math.floor(si * 2 + 2)); tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2]; } // Function to construct segment tree // from given array.This function allocates // memory for segment tree and calls // constructSTUtil() to fill the // allocated memory function constructST(arr, n) { // Fill the allocated memory st constructSTUtil(arr, 0, n - 1, 0); } let arr = [ 1, 3, 5, 7, 9, 11 ]; let n = arr.length; // Build segment tree from given array constructST(arr, n); // Print sum of values in array // from index 1 to 3 console.log(`Sum of values in given range = ${getSum(n, 1, 3)}`); // Add 10 to all nodes at indexes // from 1 to 5. updateRange(n, 1, 5, 10); console.log(`Updated sum of values in given range = ${ getSum(n, 1, 3)}`); // This code is contributed by ksam24000. OutputSum of values in given range = 15 Updated sum of values in given range = 45
Time Complexity: O(N)
Auxiliary Space: O(MAX)
Applications:
- Interval scheduling: Segment trees can be used to efficiently schedule non-overlapping intervals, such as scheduling appointments or allocating resources.
- Range-based statistics: Segment trees can be used to compute range-based statistics such as variance, standard deviation, and percentiles.
- Image processing: Segment trees are used in image processing algorithms to divide an image into segments based on color, texture, or other attributes.
Advantages:
- Efficient querying: Segment trees can be used to efficiently answer queries about the minimum, maximum, sum, or other aggregate value of a range of elements in an array.
- Efficient updates: Segment trees can be used to efficiently update a range of elements in an array, such as incrementing or decrementing a range of values.
- Flexibility: Segment trees can be used to solve a wide variety of problems involving range queries and updates.
Disadvantages:
- Complexity: Segment trees can be complex to implement and maintain, especially for large arrays or high-dimensional data.
- Time complexity: The time complexity of segment tree operations like build, update, and the query is O(log N) , which is higher than some other data structures like the Fenwick tree.
- Space complexity: The space complexity of a segment tree is O(4N) which is relatively high.
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