# Number of subarrays having sum exactly equal to k

• Difficulty Level : Medium
• Last Updated : 07 May, 2021

Given an unsorted array of integers, find the number of subarrays having sum exactly equal to a given number k.

Examples:

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```Input : arr[] = {10, 2, -2, -20, 10},
k = -10
Output : 3
Subarrays: arr[0...3], arr[1...4], arr[3..4]
have sum exactly equal to -10.

Input : arr[] = {9, 4, 20, 3, 10, 5},
k = 33
Output : 2
Subarrays : arr[0...2], arr[2...4] have 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. Following is the naive implementation –

## C++

 `// C++ program for``// the above approach``#include ``using` `namespace` `std;``int` `main()``{``  ``int` `arr[] = {10, 2, -2, -20, 10};``  ``int` `k = -10;``  ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``  ``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 Chitranayal`

## Java

 `// Java program for``// the above approach``import` `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);``    ``}``}`

## Python3

 `# Python3 program for``# the above approach``arr ``=` `[ ``10``, ``2``, ``-``2``, ``-``20``, ``10` `]``n ``=` `len``(arr)``k ``=` `-``10``res ``=` `0` `# Calculate all subarrays``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` `print``(res)` `# This code is contributed by kavan155gondalia`

## C#

 `// C# program for``// the above approach``using` `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

 ``
Output:
`3`

Efficient Solution –
An efficient solution is while traversing the array, store 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 count by the number of subarrays having sum 0 in that case.

## C++

 `// C++ program to find number of subarrays``// with sum exactly equal to k.``#include ``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);``    ``cout << findSubarraySum(arr, n, sum);``    ``return` `0;``}`

## 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 prevSum = ``new` `HashMap<>();` `        ``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.get(currsum - sum);` `            ``// Add currsum value to count of``            ``// different values of sum.``            ``Integer count = prevSum.get(currsum);``            ``if` `(count == ``null``)``                ``prevSum.put(currsum, ``1``);``            ``else``                ``prevSum.put(currsum, count + ``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

 ``
Output:
`3`

Time Complexity: O(n)
Auxiliary Space: O(n)

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