Difference Array | Range update query in O(1)
Last Updated : 05 May, 2025
You are given an integer array arr[] and a list of queries. Each query is represented as a list of integers where:
[1, l, r, x]: Adds x to all elements from arr[l] to arr[r] (inclusive).[2]: Prints the current state of the array.
You need to perform the queries in order.
Examples :
Input: arr[] = [10, 5, 20, 40], queries = [ [1, 0, 1, 10], [2], [1, 1, 3, 20], [1, 2, 2, 30], [2] ]
Output: 20 15 20 40
20 35 70 60
Explanation: [1, 0, 1, 10]: Adds 10 to arr[0] and arr[1].
Array becomes [20, 15, 20, 40].
- (2): Prints the array:
20 15 20 40. - [1, 1, 3, 20)]: Adds 20 to
arr[1], arr[2], and arr[3].
Array becomes [20, 35, 40, 60]. - [1, 2, 2, 30]: Adds 30 to
arr[2].
Array becomes [20, 35, 70, 60]. - (2): Prints the array:
20 35 70 60.
[Naive Approach] Using loops for each queries - O(n * q) time and O(1) space
A simple solution is to do following :
- update(l, r, x) : Run a loop from l to r and add x to all elements from arr[l] to arr[r]
- printArray() : Simply print arr[].
C++ #include <iostream> #include <vector> using namespace std; void update(vector<int> &arr, int l, int r, int x) { for (int i = l; i <= r; i++) { arr[i] += x; } } void printArray(const vector<int> &arr) { for (int num : arr) { cout << num << " "; } cout << endl; } int main() { vector<int> arr = {10, 5, 20, 40}; vector<vector<int>> queries = {{1, 0, 1, 10}, {2}, {1, 1, 3, 20}, {1, 2, 2, 30}, {2}}; for (const auto &query : queries) { if (query[0] == 1) { // update operation int l = query[1]; int r = query[2]; int x = query[3]; update(arr, l, r, x); } else if (query[0] == 2) { // print operation printArray(arr); } } return 0; } import java.util.Arrays; import java.util.List; public class GfG{ public static void update(int[] arr, int l, int r, int x) { for (int i = l; i <= r; i++) { arr[i] += x; } } public static void printArray(int[] arr) { for (int num : arr) { System.out.print(num + " "); } System.out.println(); } public static void main(String[] args) { int[] arr = { 10, 5, 20, 40 }; int[][] queries = { { 1, 0, 1, 10 }, { 2 }, { 1, 1, 3, 20 }, { 1, 2, 2, 30 }, { 2 } }; for (int[] query : queries) { if (query[0] == 1) { // update operation int l = query[1]; int r = query[2]; int x = query[3]; update(arr, l, r, x); } else if (query[0] == 2) { // print operation printArray(arr); } } } } def update(arr, l, r, x): for i in range(l, r + 1): arr[i] += x def print_array(arr): print(' '.join(map(str, arr))) if __name__ == '__main__': arr = [10, 5, 20, 40] queries = [[1, 0, 1, 10], [2], [1, 1, 3, 20], [1, 2, 2, 30], [2]] for query in queries: if query[0] == 1: # update operation l, r, x = query[1], query[2], query[3] update(arr, l, r, x) elif query[0] == 2: # print operation print_array(arr) using System; using System.Collections.Generic; class GfG{ static void Update(int[] arr, int l, int r, int x) { for (int i = l; i <= r; i++) { arr[i] += x; } } static void PrintArray(int[] arr) { foreach(int num in arr) { Console.Write(num + " "); } Console.WriteLine(); } static void Main() { int[] arr = new int[] { 10, 5, 20, 40 }; List<List<int>> queries = new List<List<int>>{ new List<int>{ 1, 0, 1, 10 }, new List<int>{ 2 }, new List<int>{ 1, 1, 3, 20 }, new List<int>{ 1, 2, 2, 30 }, new List<int>{ 2 } }; foreach(var query in queries) { if (query[0] == 1) { // update operation int l = query[1]; int r = query[2]; int x = query[3]; Update(arr, l, r, x); } else if (query[0] == 2) { // print operation PrintArray(arr); } } } } function update(arr, l, r, x) { for (let i = l; i <= r; i++) { arr[i] += x; } } function printArray(arr) { console.log(arr.join(" ")); } const arr = [ 10, 5, 20, 40 ]; const queries = [ [ 1, 0, 1, 10 ], [ 2 ], [ 1, 1, 3, 20 ], [ 1, 2, 2, 30 ], [ 2 ] ]; queries.forEach(query => { if (query[0] === 1) { // update operation const [_, l, r, x] = query; update(arr, l, r, x); } else if (query[0] === 2) { // print operation printArray(arr); } }); Output20 15 20 40 20 35 70 60
Time Complexity: O(n * q), O(n) for each queries
Space Complexity: O(1)
[Expected Approach] Using Difference Array
Difference array d[i] of a given array arr[i] is defined as d[i] = arr[i] - arr[i-1] (for 0 < i < n) and d[0] = arr[0] considering 0 based indexing. Difference array can be used to perform range update queries "l r x" where l is left index, r is right index and x is value to be added and after all queries you can return original array from it. Where update range operations can be performed in O(1) complexity.
- update(l, r, x) : Add x to d[l] and subtract it from d[r+1], i.e., we do d[l] += x, d[r+1] -= x
- printArray() : Do a[0] = d[0] and print it. For rest of the elements, do arr[i] = arr[i-1] + d[i] and print them.
Illustration: Let us understand this with an example arr = [2, 5, 7, 9, 6]. The difference array would be d = [2, 3, 2, 2, -3]. After an update say update(1, 3, 4), we add 4 to index 1 and subtract from index 4, the difference array would become d = [2, 7, 2, 2, -7]. Now to print array, we print 2, 2 + 7 = 9, 9 + 2 = 11, 11 + 2 = 13, 13 + (-7) = 6
C++ #include <iostream> #include <vector> using namespace std; vector<int> initDiffArray(vector<int> &arr) { int n = arr.size(); // Initialize difference array with an extra element vector<int> d(n + 1, 0); d[0] = arr[0]; for (int i = 1; i < n; i++) { d[i] = arr[i] - arr[i - 1]; } return d; } void update(vector<int> &d, int l, int r, int x) { d[l] += x; d[r + 1] -= x; } void printArray(vector<int> &arr, vector<int> &d) { for (int i = 0; i < arr.size(); i++) { if (i == 0) arr[i] = d[i]; else arr[i] = d[i] + arr[i - 1]; cout << arr[i] << " "; } cout << endl; } int main() { vector<int> arr{10, 5, 20, 40}; vector<int> d = initDiffArray(arr); vector<vector<int>> queries = {{1, 0, 1, 10}, {2}, {1, 1, 3, 20}, {1, 2, 2, 30}, {2}}; for (const auto &query : queries) { if (query[0] == 1) { // If it's an update query: update(l, r, x) int l = query[1]; int r = query[2]; int x = query[3]; update(d, l, r, x); } else if (query[0] == 2) { // If it's a print query: printArray() printArray(arr, d); } } return 0; } import java.util.Arrays; import java.util.List; import java.util.ArrayList; public class GfG{ public static int[] initDiffArray(int[] arr) { int n = arr.length; int[] d = new int[n + 1]; d[0] = arr[0]; for (int i = 1; i < n; i++) { d[i] = arr[i] - arr[i - 1]; } return d; } public static void update(int[] d, int l, int r, int x) { d[l] += x; d[r + 1] -= x; } public static void printArray(int[] arr, int[] d) { for (int i = 0; i < arr.length; i++) { if (i == 0) arr[i] = d[i]; else arr[i] = d[i] + arr[i - 1]; System.out.print(arr[i] + " "); } System.out.println(); } public static void main(String[] args) { int[] arr = {10, 5, 20, 40}; int[] d = initDiffArray(arr); List<int[]> queries = new ArrayList<>(); queries.add(new int[]{1, 0, 1, 10}); queries.add(new int[]{2}); queries.add(new int[]{1, 1, 3, 20}); queries.add(new int[]{1, 2, 2, 30}); queries.add(new int[]{2}); for (int[] query : queries) { if (query[0] == 1) { // If it's an update query: update(l, r, x) int l = query[1]; int r = query[2]; int x = query[3]; update(d, l, r, x); } else if (query[0] == 2) { // If it's a print query: printArray() printArray(arr, d); } } } } def init_diff_array(arr): n = len(arr) d = [0] * (n + 1) d[0] = arr[0] for i in range(1, n): d[i] = arr[i] - arr[i - 1] return d def update(d, l, r, x): d[l] += x d[r + 1] -= x def print_array(arr, d): for i in range(len(arr)): if i == 0: arr[i] = d[i] else: arr[i] = d[i] + arr[i - 1] print(arr[i], end=' ') print() if __name__ == '__main__': arr = [10, 5, 20, 40] d = init_diff_array(arr) queries = [[1, 0, 1, 10], [2], [1, 1, 3, 20], [1, 2, 2, 30], [2]] for query in queries: if query[0] == 1: # If it's an update query: update(l, r, x) l, r, x = query[1], query[2], query[3] update(d, l, r, x) elif query[0] == 2: # If it's a print query: print_array() print_array(arr, d)
using System; using System.Collections.Generic; class GfG{ static int[] InitDiffArray(int[] arr) { int n = arr.Length; int[] d = new int[n + 1]; d[0] = arr[0]; for (int i = 1; i < n; i++) { d[i] = arr[i] - arr[i - 1]; } return d; } static void Update(int[] d, int l, int r, int x) { d[l] += x; d[r + 1] -= x; } static void PrintArray(int[] arr, int[] d) { for (int i = 0; i < arr.Length; i++) { if (i == 0) arr[i] = d[i]; else arr[i] = d[i] + arr[i - 1]; Console.Write(arr[i] + " "); } Console.WriteLine(); } static void Main() { int[] arr = { 10, 5, 20, 40 }; int[] d = InitDiffArray(arr); List<int[]> queries = new List<int[]>{ new int[] { 1, 0, 1, 10 }, new int[] { 2 }, new int[] { 1, 1, 3, 20 }, new int[] { 1, 2, 2, 30 }, new int[] { 2 } }; foreach(var query in queries) { if (query[0] == 1) { // If it's an update query: update(l, r, x) int l = query[1]; int r = query[2]; int x = query[3]; Update(d, l, r, x); } else if (query[0] == 2) { // If it's a print query: printArray() PrintArray(arr, d); } } } } function initDiffArray(arr) { const n = arr.length; const d = new Array(n + 1).fill(0); d[0] = arr[0]; for (let i = 1; i < n; i++) { d[i] = arr[i] - arr[i - 1]; } return d; } function update(d, l, r, x) { d[l] += x; d[r + 1] -= x; } function printArray(arr, d) { for (let i = 0; i < arr.length; i++) { if (i === 0) arr[i] = d[i]; else arr[i] = d[i] + arr[i - 1]; process.stdout.write(arr[i] + " "); } console.log(); } const arr = [ 10, 5, 20, 40 ]; const d = initDiffArray(arr); const queries = [ [ 1, 0, 1, 10 ], [ 2 ], [ 1, 1, 3, 20 ], [ 1, 2, 2, 30 ], [ 2 ] ]; queries.forEach(query => { if (query[0] === 1) { // If it's an update query: update(l, r, x) const l = query[1]; const r = query[2]; const x = query[3]; update(d, l, r, x); } else if (query[0] === 2) { // If it's a print query: printArray() printArray(arr, d); } }); Output:
20 15 20 40 20 35 70 60
Time complexity:
For update here is improved to O(1).
For printArray() still takes O(n) time.
Auxiliary Space: O(n)
Similar Reads
PreComputation Technique on Arrays Precomputation refers to the process of pre-calculating and storing the results of certain computations or data structures(array in this case) in advance, in order to speed up the execution time of a program. This can be useful in situations where the same calculations are needed multiple times, as
15 min read
Queries for the product of first N factorials Given Q[] queries where each query consists of an integer N, the task is to find the product of first N factorials for each of the query. Since the result could be large, compute it modulo 109 + 7.Examples: Input: Q[] = {4, 5} Output: 288 34560 Query 1: 1! * 2! * 3! * 4! = 1 * 2 * 6 * 24 = 288 Query
7 min read
Range sum queries without updates Given an array arr of integers of size n. We need to compute the sum of elements from index i to index j. The queries consisting of i and j index values will be executed multiple times.Examples: Input : arr[] = {1, 2, 3, 4, 5} i = 1, j = 3 i = 2, j = 4Output : 9 12 Input : arr[] = {1, 2, 3, 4, 5} i
6 min read
Range Queries for Frequencies of array elements Given an array of n non-negative integers. The task is to find frequency of a particular element in the arbitrary range of array[]. The range is given as positions (not 0 based indexes) in array. There can be multiple queries of given type. Examples: Input : arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11}; lef
13 min read
Count Primes in Ranges Given a 2d array queries[][] of size n, where each query queries[i] contain 2 elements [l, r], your task is to find the count of number of primes in inclusive range [l, r]Examples: Input: queries[][] = [ [1, 10], [5, 10], [11, 20] ]Output: 4 2 4Explanation: For query 1, number of primes in range [1,
12 min read
Check in binary array the number represented by a subarray is odd or even Given an array such that all its terms is either 0 or 1.You need to tell the number represented by a subarray a[l..r] is odd or even Examples : Input : arr = {1, 1, 0, 1} l = 1, r = 3 Output : odd number represented by arr[l...r] is 101 which 5 in decimal form which is odd Input : arr = {1, 1, 1, 1}
4 min read
GCDs of given index ranges in an Array Given an array arr[] of size N and Q queries of type {qs, qe} where qs and qe denote the starting and ending index of the query, the task is to find the GCD of all the numbers in the range. Examples: Input: arr[] = {2, 3, 60, 90, 50};Index Ranges: {1, 3}, {2, 4}, {0, 2}Output: GCDs of given ranges a
14 min read
Mean of range in array Given an array arr[] of n integers and q queries represented by an array queries[][], where queries[i][0] = l and queries[i][1] = r. For each query, the task is to calculate the mean of elements in the range l to r and return its floor value. Examples: Input: arr[] = [3, 7, 2, 8, 5] queries[][] = [[
12 min read
Difference Array | Range update query in O(1) You are given an integer array arr[] and a list of queries. Each query is represented as a list of integers where:[1, l, r, x]: Adds x to all elements from arr[l] to arr[r] (inclusive).[2]: Prints the current state of the array.You need to perform the queries in order.Examples : Input: arr[] = [10,
10 min read
Range sum query using Sparse Table We have an array arr[]. We need to find the sum of all the elements in the range L and R where 0 <= L <= R <= n-1. Consider a situation when there are many range queries. Examples: Input : 3 7 2 5 8 9 query(0, 5) query(3, 5) query(2, 4) Output : 34 22 15Note : array is 0 based indexed and q
8 min read