Reconstructing Segment Tree

Last Updated : 09 Mar, 2023

We are given 2*N - 1 integers. We need to check whether it is possible to construct a Range Minimum Query segment tree for an array of N distinct integers from these integers. If so, we must output the segment tree array. N is given to be a power of 2.
An RMQ segment tree is a binary tree where each node is equal to the minimum value of its children. This type of tree is used to efficiently find the minimum value of elements in a given range.

Input  : 1 1 1 1 2 2 3 3 3 4 4 5 6 7 8 Output : 1 1 3 1 2 3 4 1 5 2 6 3 7 4 8 The segment tree is shown below    Input  : -381 -460 -381 95 -460 855 -242           405 -460 982 -381 -460 95 981 855 Output : -460 -460 -381 -460 95 -381 855          -460 -242 95 405 -381 981 855 982  By constructing a segment tree from the output, we can see that it a valid tree for RMQ and the leaves are all distinct integers.

What we first do is iterate through the given integers counting the number of occurrences of each number, then sorting them by value. In C++, we can use the data structure map, which stores elements in sorted order.
Now we maintain a queue for each possible level of the segment tree. We put the initial root of the tree (array index 0) into the queue for the max level. We then insert the smallest element into the leftmost nodes. We then detach these nodes from the main tree. As we detach a node, we create a new tree of height h - 1, where h is the height of the current node. We can see this in figure 2. We insert the root node of this new tree into the appropriate queue based on its height.
We go through each element, getting a tree of the appropriate height based on the number of occurrences of that element. If at any point such a tree does not exist, then it is not possible to create a segment tree.

CPP

// C++ Program to Create RMQ Segment Tree #include <bits/stdc++.h> using namespace std;  // Returns true if it is possible to construct // a range minimum segment tree from given array. bool createTree(int arr[], int N) {     // Store the height of the final tree     const int height = log2(N) + 1;      // Container to sort and store occurrences of elements     map<int, int> multi;      // Insert elements into the container     for (int i = 0; i < 2 * N - 1; ++i)         ++multi[arr[i]];      // Used to store new subtrees created     set<int> Q[height];      // Insert root into set     Q[height - 1].emplace(0);      // Iterate through each unique element in set     for (map<int, int>::iterator it = multi.begin();         it != multi.end(); ++it)     {         // Number of occurrences is greater than height         // Or, no subtree exists that can accommodate it         if (it->second > height || Q[it->second - 1].empty())             return false;          // Get the appropriate subtree         int node = *Q[it->second - 1].begin();          // Delete the subtree we grabbed         Q[it->second - 1].erase(Q[it->second - 1].begin());          int level = 1;         for (int i = node; i < 2 * N - 1;             i = 2 * i + 1, ++level)         {             // Insert new subtree created into root             if (2 * i + 2 < 2 * N - 1)                 Q[it->second - level - 1].emplace(2 * i + 2);              // Insert element into array at position             arr[i] = it->first;         }     }     return true; }  // Driver program int main() {     int N = 8;     int arr[2 * N - 1] = {1, 1, 1, 1, 2, 2,                  3, 3, 3, 4, 4, 5, 6, 7, 8};     if (createTree(arr, N))     {         cout << "YES\n";         for (int i = 0; i < 2 * N - 1; ++i)             cout << arr[i] << " ";     }     else         cout << "NO\n";     return 0; }

Java

import java.util.*;  public class RMQSegmentTree {      // Returns true if it is possible to construct     // a range minimum segment tree from given array.     public static boolean createTree(int[] arr, int N)     {         // Store the height of the final tree         final int height             = (int)(Math.log(N) / Math.log(2)) + 1;          // Container to sort and store occurrences of         // elements         Map<Integer, Integer> multi = new HashMap<>();          // Insert elements into the container         for (int i = 0; i < 2 * N - 1; ++i)             multi.put(arr[i],                       multi.getOrDefault(arr[i], 0) + 1);          // Used to store new subtrees created         Set<Integer>[] Q = new HashSet[height];         for (int i = 0; i < height; i++) {             Q[i] = new HashSet<Integer>();         }          // Insert root into set         Q[height - 1].add(0);          // Iterate through each unique element in set         for (Map.Entry<Integer, Integer> entry :              multi.entrySet()) {             int key = entry.getKey();             int value = entry.getValue();              // Number of occurrences is greater than height             // Or, no subtree exists that can accommodate it             if (value > height || Q[value - 1].isEmpty())                 return false;              // Get the appropriate subtree             int node = Q[value - 1].iterator().next();              // Delete the subtree we grabbed             Q[value - 1].remove(node);              int level = 1;             for (int i = node; i < 2 * N - 1;                  i = 2 * i + 1, ++level) {                 // Insert new subtree created into root                 if (2 * i + 2 < 2 * N - 1)                     Q[value - level - 1].add(2 * i + 2);                  // Insert element into array at position                 arr[i] = key;             }         }         return true;     }      // Driver program     public static void main(String[] args)     {         int N = 8;         int[] arr = { 1, 1, 1, 1, 2, 2, 3, 3,                       3, 4, 4, 5, 6, 7, 8 };          if (createTree(arr, N)) {             System.out.println("YES");             for (int i = 0; i < 2 * N - 1; ++i)                 System.out.print(arr[i] + " ");         }         else {             System.out.println("NO");         }     } }

Python3

# Python Program to Create RMQ Segment Tree from typing import List from math import log2  # Returns true if it is possible to construct # a range minimum segment tree from given array. def createTree(arr: List[int], N: int) -> bool:      # Store the height of the final tree     height = int(log2(N)) + 1      # Container to sort and store occurrences of elements     multi = {}      # Insert elements into the container     for i in range(2 * N - 1):         if arr[i] not in multi:             multi[arr[i]] = 0         multi[arr[i]] += 1      # Used to store new subtrees created     Q = [set() for _ in range(height)]      # Insert root into set     Q[height - 1].add(0)      # Iterate through each unique element in set     for k, v in multi.items():          # Number of occurrences is greater than height         # Or, no subtree exists that can accommodate it         if (v > height or len(Q[v - 1]) == 0):             return False          # Get the appropriate subtree         node = sorted(Q[v - 1])[0]          # Delete the subtree we grabbed         Q[v - 1].remove(sorted(Q[v - 1])[0])         level = 1                  # for (int i = node; i < 2 * N - 1;         #     i = 2 * i + 1, ++level)         i = node         while i < 2 * N - 1:              # Insert new subtree created into root             if (2 * i + 2 < 2 * N - 1):                 Q[v - level - 1].add(2 * i + 2)              # Insert element into array at position             arr[i] = k             level += 1             i = 2 * i + 1     return True   # Driver program if __name__ == "__main__":      N = 8     arr = [1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 6, 7, 8]     if (createTree(arr, N)):          print("YES")         # for (int i = 0; i < 2 * N - 1; ++i)         for i in range(2 * N - 1):             print(arr[i], end=" ")      else:         print("No")  # This code is contributed by sanjeev2552

// C# Program to Create RMQ Segment Tree using System; using System.Collections.Generic;  public class RMQSegmentTree {          // Returns true if it is possible to construct     // a range minimum segment tree from given array.     static bool CreateTree(int[] arr, int N)     {         // Store the height of the final tree         int height = (int)(Math.Log(N) / Math.Log(2)) + 1;          // Container to sort and store occurrences of         // elements         Dictionary<int, int> multi             = new Dictionary<int, int>();          // Insert elements into the container         for (int i = 0; i < 2 * N - 1; ++i)             if (multi.ContainsKey(arr[i]))                 multi[arr[i]]++;             else                 multi[arr[i]] = 1;          // Used to store new subtrees created         SortedSet<int>[] Q = new SortedSet<int>[ height ];          // Initialize each set in the array         for (int i = 0; i < height; i++)             Q[i] = new SortedSet<int>();          // Insert root into set         Q[height - 1].Add(0);          // Iterate through each unique element in set         foreach(KeyValuePair<int, int> entry in multi)         {             int count = entry.Value;             int key = entry.Key;              // Number of occurrences is greater than height             // Or, no subtree exists that can accommodate it             if (count > height || Q[count - 1].Count == 0)                 return false;              // Get the appropriate subtree             int node = Q[count - 1].Min;              // Delete the subtree we grabbed             Q[count - 1].Remove(node);              int level = 1;             for (int i = node; i < 2 * N - 1;                  i = 2 * i + 1, ++level) {                                       // Insert new subtree created into root                 if (2 * i + 2 < 2 * N - 1)                     Q[count - level - 1].Add(2 * i + 2);                  // Insert element into array at position                 arr[i] = key;             }         }         return true;     }      // Driver program     static public void Main()     {         int N = 8;         int[] arr = { 1, 1, 1, 1, 2, 2, 3, 3,                       3, 4, 4, 5, 6, 7, 8 };          if (CreateTree(arr, N)) {             Console.WriteLine("YES");             for (int i = 0; i < 2 * N - 1; ++i)                 Console.Write(arr[i] + " ");         }         else             Console.WriteLine("NO");          Console.ReadLine();     } } // This code is contributed by prasad264

JavaScript

// JavaScript Program to Create RMQ Segment Tree  // Returns true if it is possible to construct // a range minimum segment tree from given array. function createTree(arr, N) {  // Store the height of the final tree const height = Math.floor(Math.log2(N)) + 1;  // Container to sort and store occurrences of elements const multi = {};  // Insert elements into the container for (let i = 0; i < 2 * N - 1; i++) { if (multi[arr[i]] == undefined) { multi[arr[i]] = 0; } multi[arr[i]] += 1; }  // Used to store new subtrees created const Q = new Array(height); for (let i = 0; i < height; i++) { Q[i] = new Set(); }  // Insert root into set Q[height - 1].add(0);  // Iterate through each unique element in set for (const [k, v] of Object.entries(multi)) {// Number of occurrences is greater than height // Or, no subtree exists that can accommodate it if (v > height || Q[v - 1].size == 0) {   return false; }  // Get the appropriate subtree const node = Math.min(...Q[v - 1]);  // Delete the subtree we grabbed Q[v - 1].delete(node); let level = 1; let i = node;  while (i < 2 * N - 1) {    // Insert new subtree created into root   if (2 * i + 2 < 2 * N - 1) {     Q[v - level - 1].add(2 * i + 2);   }    // Insert element into array at position   arr[i] = k;   level += 1;   i = 2 * i + 1; } } return true; }  // Driver program const N = 8; const arr = [1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 6, 7, 8]; if (createTree(arr, N)) { console.log("YES"); let ans=""; for (let i = 0; i < 2 * N - 1; i++) { ans = ans + arr[i] + " "; } console.log(ans); } else { console.log("No"); }

Output:

YES 1 1 3 1 2 3 4 1 5 2 6 3 7 4 8

Main time complexity is caused by sorting elements.
Time Complexity: O(N log N)
Space Complexity: O(N)

Reconstructing Segment Tree

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