Given an unsorted array of integers, find the number of subarrays having a sum exactly equal to a given number k.
Examples:
Input : arr[] = {10, 2, -2, -20, 10}, k = -10
Output : 3
Explanation: Subarrays: arr[0…3], arr[1…4], arr[3..4] have a sum exactly equal to -10.Input : arr[] = {9, 4, 20, 3, 10, 5}, k = 33
Output : 2
Explanation: Subarrays : arr[0…2], arr[2…4] have a sum exactly equal to 33.
Naive Solution: A simple solution is to traverse all the subarrays and calculate their sum. If the sum is equal to the required sum, then increment the count of subarrays. Print final count of subarray.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;int main(){ int arr[] = {10, 2, -2, -20, 10}; int k = -10; int n = sizeof(arr) / sizeof(arr[0]); int res = 0; // Calculate all subarrays for (int i = 0; i < n; i++) { int sum = 0; for (int j = i; j < n; j++) { // Calculate required sum sum += arr[j]; // Check if sum is equal to required sum if (sum == k) res++; } } cout << (res) << endl;}// This code is contributed by Aditya Kumar (adityakumar129) |
C
// C program for the above approach#include <stdio.h>int main(){ int arr[] = { 10, 2, -2, -20, 10 }; int k = -10; int n = sizeof(arr) / sizeof(arr[0]); int res = 0; // Calculate all subarrays for (int i = 0; i < n; i++) { int sum = 0; for (int j = i; j < n; j++) { // Calculate required sum sum += arr[j]; // Check if sum is equal to required sum if (sum == k) res++; } } printf("%d\n", res);}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java program for the above approachimport java.util.*;class Solution { public static void main(String[] args) { int arr[] = { 10, 2, -2, -20, 10 }; int k = -10; int n = arr.length; int res = 0; // calculate all subarrays for (int i = 0; i < n; i++) { int sum = 0; for (int j = i; j < n; j++) { // calculate required sum sum += arr[j]; // check if sum is equal to required sum if (sum == k) res++; } } System.out.println(res); }}// This code is contributed by Aditya Kumar (adityakumar129) |
Python3
# Python3 program for# the above approach# Calculate all subarraysdef count_all_subarrys(arr, n): # count all subarrays res = 0 for i in range(n): summ = 0 for j in range(i, n): # Calculate required sum summ += arr[j] # Check if sum is equal to # required sum if summ == k: res += 1 return res# main functionif __name__ == "__main__": arr = [10, 2, -2, -20, 10] n = len(arr) k = -10 print(count_all_subarrys(arr, n))# This code is contributed by shushant kumar |
C#
// C# program for// the above approachusing System;using System.Collections.Generic; class GFG { static void Main() { int[] arr = {10, 2, -2, -20, 10}; int k = -10; int n = arr.Length; int res = 0; // Calculate all subarrays for (int i = 0; i < n; i++) { int sum = 0; for (int j = i; j < n; j++) { // Calculate required sum sum += arr[j]; // Check if sum is equal to // required sum if (sum == k) res++; } } Console.WriteLine(res); }}// This code is contributed by divyesh072019 |
Javascript
<script> // Javascript program for// the above approachlet arr = [ 10, 2, -2, -20, 10 ];let k = -10;let n = arr.length;let res = 0;// Calculate all subarraysfor(let i = 0; i < n; i++) { let sum = 0; for(let j = i; j < n; j++) { // Calculate required sum sum += arr[j]; // Check if sum is equal to // required sum if (sum == k) res++; }}document.write(res);// This code is contributed by suresh07 </script> |
Output
3
Time Complexity : O(n2)
Auxiliary Space: O(1)
Efficient Solution :
An efficient solution is while traversing the array, storing sum so far in currsum. Also, maintain the count of different values of currsum in a map. If the value of currsum is equal to the desired sum at any instance increment count of subarrays by one.
The value of currsum exceeds the desired sum by currsum – sum. If this value is removed from currsum then the desired sum can be obtained. From the map, find the number of subarrays previously found having sum equal to currsum-sum. Excluding all those subarrays from the current subarray, gives new subarrays having the desired sum.
So increase count by the number of such subarrays. Note that when currsum is equal to the desired sum then also check the number of subarrays previously having a sum equal to 0. Excluding those subarrays from the current subarray gives new subarrays having the desired sum. Increase the count by the number of subarrays having sum 0 in that case.
Implementation:
C++
// C++ program to find number of subarrays with sum exactly// equal to k.#include <bits/stdc++.h>using namespace std;// Function to find number of subarrays with sum exactly// equal to k.int findSubarraySum(int arr[], int n, int sum){ // STL map to store number of subarrays starting from // index zero having particular value of sum. unordered_map<int, int> prevSum; int res = 0; // Sum of elements so far. int currSum = 0; for (int i = 0; i < n; i++) { // Add current element to sum so far. currSum += arr[i]; // If currsum is equal to desired sum, then a new // subarray is found. So increase count of // subarrays. if (currSum == sum) res++; // currsum exceeds given sum by currsum - sum. Find // number of subarrays having this sum and exclude // those subarrays from currsum by increasing count // by same amount. if (prevSum.find(currSum - sum) != prevSum.end()) res += (prevSum[currSum - sum]); // Add currsum value to count of different values of // sum. prevSum[currSum]++; } return res;}int main(){ int arr[] = { 10, 2, -2, -20, 10 }; int sum = -10; int n = sizeof(arr) / sizeof(arr[0]); cout << findSubarraySum(arr, n, sum); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java program to find number of subarrays// with sum exactly equal to k.import java.util.HashMap;import java.util.Map;public class GfG { // Function to find number of subarrays // with sum exactly equal to k. static int findSubarraySum(int arr[], int n, int sum) { // HashMap to store number of subarrays // starting from index zero having // particular value of sum. HashMap<Integer, Integer> prevSum = new HashMap<>(); prevSum.put(0,1); int res = 0; // Sum of elements so far. int currSum = 0; for (int i = 0; i < n; i++) { // Add current element to sum so far. currSum += arr[i]; //calculate the sum that have to be removed //so that we can get the desired sum int removeSum=currSum-sum; //get count of occurrences of that sum that //have to removed and add it to res value if (prevSum.containsKey(removeSum)) res += prevSum.get(removeSum); // Add currsum value to count of // different values of sum. prevSum.put(currSum,prevSum.getOrDefault(currSum,0)+1); } return res; } public static void main(String[] args) { int arr[] = { 10, 2, -2, -20, 10 }; int sum = -10; int n = arr.length; System.out.println(findSubarraySum(arr, n, sum)); }}// This code is contributed by Rituraj Jain |
Python3
# Python3 program to find the number of # subarrays with sum exactly equal to k. from collections import defaultdict# Function to find number of subarrays # with sum exactly equal to k. def findSubarraySum(arr, n, Sum): # Dictionary to store number of subarrays # starting from index zero having # particular value of sum. prevSum = defaultdict(lambda : 0) res = 0 # Sum of elements so far. currsum = 0 for i in range(0, n): # Add current element to sum so far. currsum += arr[i] # If currsum is equal to desired sum, # then a new subarray is found. So # increase count of subarrays. if currsum == Sum: res += 1 # currsum exceeds given sum by currsum - sum. # Find number of subarrays having # this sum and exclude those subarrays # from currsum by increasing count by # same amount. if (currsum - Sum) in prevSum: res += prevSum[currsum - Sum] # Add currsum value to count of # different values of sum. prevSum[currsum] += 1 return res if __name__ == "__main__": arr = [10, 2, -2, -20, 10] Sum = -10 n = len(arr) print(findSubarraySum(arr, n, Sum)) # This code is contributed by Rituraj Jain |
C#
// C# program to find number of subarrays// with sum exactly equal to k.using System;using System.Collections.Generic;class GFG { // Function to find number of subarrays // with sum exactly equal to k. public static int findSubarraySum(int[] arr, int n, int sum) { // HashMap to store number of subarrays // starting from index zero having // particular value of sum. Dictionary<int, int> prevSum = new Dictionary<int, int>(); int res = 0; // Sum of elements so far int currsum = 0; for (int i = 0; i < n; i++) { // Add current element to sum so far. currsum += arr[i]; // If currsum is equal to desired sum, // then a new subarray is found. So // increase count of subarrays. if (currsum == sum) res++; // currsum exceeds given sum by currsum // - sum. Find number of subarrays having // this sum and exclude those subarrays // from currsum by increasing count by // same amount. if (prevSum.ContainsKey(currsum - sum)) res += prevSum[currsum - sum]; // Add currsum value to count of // different values of sum. if (!prevSum.ContainsKey(currsum)) prevSum.Add(currsum, 1); else { int count = prevSum[currsum]; prevSum[currsum] = count + 1; } } return res; } // Driver Code public static void Main() { int[] arr = { 10, 2, -2, -20, 10 }; int sum = -10; int n = arr.Length; Console.Write(findSubarraySum(arr, n, sum)); }}// This code is contributed by// sanjeev2552 |
Javascript
<script>// Javascript program to find number of subarrays// with sum exactly equal to k. // Function to find number of subarrays // with sum exactly equal to k. function findSubarraySum(arr,n,sum) { // HashMap to store number of subarrays // starting from index zero having // particular value of sum. let prevSum = new Map(); let res = 0; // Sum of elements so far. let currsum = 0; for (let i = 0; i < n; i++) { // Add current element to sum so far. currsum += arr[i]; // If currsum is equal to desired sum, // then a new subarray is found. So // increase count of subarrays. if (currsum == sum) res++; // currsum exceeds given sum by currsum // - sum. Find number of subarrays having // this sum and exclude those subarrays // from currsum by increasing count by // same amount. if (prevSum.has(currsum - sum)) res += prevSum.get(currsum - sum); // Add currsum value to count of // different values of sum. let count = prevSum.get(currsum); if (count == null) prevSum.set(currsum, 1); else prevSum.set(currsum, count + 1); } return res; } let arr = [10, 2, -2, -20, 10]; let sum = -10; let n = arr.length; document.write(findSubarraySum(arr, n, sum)); // This code is contributed by avanitrachhadiya2155.</script> |
Output
3
Time Complexity: O(n)
Auxiliary Space: O(n)
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