Subarray of size k with given sum
Last Updated : 01 Sep, 2022
Given an array arr[], an integer K and a Sum. The task is to check if there exists any subarray with K elements whose sum is equal to the given sum. If any of the subarray with size K has the sum equal to the given sum then print YES otherwise print NO.
Examples:
Input: arr[] = {1, 4, 2, 10, 2, 3, 1, 0, 20} k = 4, sum = 18 Output: YES Subarray = {4, 2, 10, 2} Input: arr[] = {1, 4, 2, 10, 2, 3, 1, 0, 20} k = 3, sum = 6 Output: YES A simple solution is to use nested loops, where we check every subarray of size k.
Below is the implementation of the above approach:
C++ // CPP program to check if any Subarray of size // K has a given Sum #include <iostream> using namespace std; // Function to check if any Subarray of size K // has a given Sum bool checkSubarraySum(int arr[], int n, int k, int sum) { // Consider all blocks starting with i. for (int i = 0; i < n - k + 1; i++) { int current_sum = 0; // Consider each subarray of size k for (int j = 0; j < k; j++) current_sum = current_sum + arr[i + j]; if (current_sum == sum) return true; } return false; } // Driver code int main() { int arr[] = { 1, 4, 2, 10, 2, 3, 1, 0, 20 }; int k = 4; int sum = 18; int n = sizeof(arr) / sizeof(arr[0]); if (checkSubarraySum(arr, n, k, sum)) cout << "YES"; else cout << "NO"; return 0; } // Java program to check // if any Subarray of size // K has a given Sum class GFG { // Function to check if any // Subarray of size K has // a given Sum static boolean checkSubarraySum(int arr[], int n, int k, int sum) { // Consider all blocks // starting with i. for (int i = 0; i < n - k + 1; i++) { int current_sum = 0; // Consider each // subarray of size k for (int j = 0; j < k; j++) current_sum = current_sum + arr[i + j]; if (current_sum == sum) return true; } return false; } // Driver code public static void main(String args[]) { int arr[] = new int[] { 1, 4, 2, 10, 2, 3, 1, 0, 20 }; int k = 4; int sum = 18; int n = arr.length; if (checkSubarraySum(arr, n, k, sum)) System.out.println("YES"); else System.out.println("NO"); } } // This code is contributed // by Kirti_Mangal # Python3 program to check # if any Subarray of size # K has a given Sum # Function to check if # any Subarray of size # K, has a given Sum def checkSubarraySum(arr, n, k, sum): # Consider all blocks # starting with i. for i in range (n - k + 1): current_sum = 0; # Consider each subarray # of size k for j in range(k): current_sum = (current_sum + arr[i + j]); if (current_sum == sum): return True; return False; # Driver code arr = [1, 4, 2, 10, 2, 3, 1, 0, 20]; k = 4; sum = 18; n = len(arr); if (checkSubarraySum(arr, n, k, sum)): print("YES"); else: print("NO"); # This code is contributed by mits // C# program to check if any // Subarray of size K has a given Sum using System; class GFG { // Function to check if any Subarray // of size K has a given Sum static bool checkSubarraySum(int[] arr, int n, int k, int sum) { // Consider all blocks // starting with i. for (int i = 0; i < n - k + 1; i++) { int current_sum = 0; // Consider each // subarray of size k for (int j = 0; j < k; j++) current_sum = current_sum + arr[i + j]; if (current_sum == sum) return true; } return false; } // Driver code static void Main() { int[] arr = new int[] { 1, 4, 2, 10, 2, 3, 1, 0, 20 }; int k = 4; int sum = 18; int n = arr.Length; if (checkSubarraySum(arr, n, k, sum)) Console.WriteLine("YES"); else Console.WriteLine("NO"); } } // This code is contributed // by mits <?php // PHP program to check // if any Subarray of size // K has a given Sum // Function to check if // any Subarray of size // K, has a given Sum function checkSubarraySum($arr, $n, $k, $sum) { // Consider all blocks starting with i. for ($i = 0; $i < $n - $k + 1; $i++) { $current_sum = 0; // Consider each subarray of size k for ($j = 0; $j < $k; $j++) $current_sum = $current_sum + $arr[$i + $j]; if ($current_sum == $sum) return true; } return false; } // Driver code $arr = array(1, 4, 2, 10, 2, 3, 1, 0, 20); $k = 4; $sum = 18; $n = sizeof($arr); if (checkSubarraySum($arr, $n, $k, $sum)) echo "YES"; else echo "NO"; // This code is contributed by mits ?> <script> // Javascript program to check if any // Subarray of size K has a given Sum // Function to check if any Subarray // of size K has a given Sum function checkSubarraySum(arr, n, k, sum) { // Consider all blocks // starting with i. for (let i = 0; i < n - k + 1; i++) { let current_sum = 0; // Consider each // subarray of size k for (let j = 0; j < k; j++) current_sum = current_sum + arr[i + j]; if (current_sum == sum) return true; } return false; } let arr = [ 1, 4, 2, 10, 2, 3, 1, 0, 20 ]; let k = 4; let sum = 18; let n = arr.length; if (checkSubarraySum(arr, n, k, sum)) document.write("YES"); else document.write("NO"); // This code is contributed by mukesh07. </script> Time Complexity: O(n * k)
An efficient solution is to check sliding window technique and simultaneously check if the sum is equal to the given sum.
Implementation:
C++ // CPP program to check if any Subarray of size // K has a given Sum #include <iostream> using namespace std; // Function to check if any Subarray of size K // has a given Sum bool checkSubarraySum(int arr[], int n, int k, int sum) { // Check for first window int curr_sum = 0; for (int i=0; i<k; i++) curr_sum += arr[i]; if (curr_sum == sum) return true; // Consider remaining blocks ending with j for (int j = k; j < n; j++) { curr_sum = curr_sum + arr[j] - arr[j-k]; if (curr_sum == sum) return true; } return false; } // Driver code int main() { int arr[] = { 1, 4, 2, 10, 2, 3, 1, 0, 20 }; int k = 4; int sum = 18; int n = sizeof(arr) / sizeof(arr[0]); if (checkSubarraySum(arr, n, k, sum)) cout << "YES"; else cout << "NO"; return 0; } // Java program to check if any Subarray of size // K has a given Sum class GFG{ // Function to check if any Subarray of size K // has a given Sum static boolean checkSubarraySum(int[] arr, int n, int k, int sum) { // Check for first window int curr_sum = 0; for (int i=0; i<k; i++) curr_sum += arr[i]; if (curr_sum == sum) return true; // Consider remaining blocks ending with j for (int j = k; j < n; j++) { curr_sum = curr_sum + arr[j] - arr[j-k]; if (curr_sum == sum) return true; } return false; } // Driver code public static void main(String[] args) { int[] arr=new int[]{ 1, 4, 2, 10, 2, 3, 1, 0, 20 }; int k = 4; int sum = 18; int n = arr.length; if (checkSubarraySum(arr, n, k, sum)) System.out.println("YES"); else System.out.println("NO"); } } // This code is contributed by mits # Python3 program to check if any # Subarray of size K has a given Sum # Function to check if any Subarray # of size K has a given Sum def checkSubarraySum(arr, n, k, sumV): # Check for first window curr_sum = 0 for i in range(0, k): curr_sum += arr[i] if (curr_sum == sumV): return true # Consider remaining blocks # ending with j for j in range(k, n): curr_sum = (curr_sum + arr[j] - arr[j - k]) if (curr_sum == sumV) : return True return False # Driver code arr = [ 1, 4, 2, 10, 2, 3, 1, 0, 20 ] k = 4 sumV = 18 n = len(arr) if (checkSubarraySum(arr, n, k, sumV)): print("YES") else: print( "NO") # This code is contributed # by Yatin Gupta // C# program to check if any Subarray of size // K has a given Sum using System; class GFG{ // Function to check if any Subarray of size K // has a given Sum static bool checkSubarraySum(int[] arr, int n, int k, int sum) { // Check for first window int curr_sum = 0; for (int i=0; i<k; i++) curr_sum += arr[i]; if (curr_sum == sum) return true; // Consider remaining blocks ending with j for (int j = k; j < n; j++) { curr_sum = curr_sum + arr[j] - arr[j-k]; if (curr_sum == sum) return true; } return false; } // Driver code static void Main() { int[] arr=new int[]{ 1, 4, 2, 10, 2, 3, 1, 0, 20 }; int k = 4; int sum = 18; int n = arr.Length; if (checkSubarraySum(arr, n, k, sum)) Console.WriteLine("YES"); else Console.WriteLine("NO"); } } // This code is contributed by mits <?php // PHP program to check if any // Subarray of size K has a given Sum // Function to check if any Subarray // of size K has a given Sum function checkSubarraySum($arr, $n, $k, $sum) { // Check for first window $curr_sum = 0; for ($i = 0; $i < $k; $i++) $curr_sum += $arr[$i]; if ($curr_sum == $sum) return true; // Consider remaining blocks // ending with j for ($j = $k; $j < $n; $j++) { $curr_sum = $curr_sum + $arr[$j] - $arr[$j - $k]; if ($curr_sum == $sum) return true; } return false; } // Driver code $arr = array( 1, 4, 2, 10, 2, 3, 1, 0, 20 ); $k = 4; $sum = 18; $n = count($arr); if (checkSubarraySum($arr, $n, $k, $sum)) echo "YES"; else echo "NO"; // This code is contributed // by inder_verma ?> <script> // Javascript program to check if any // Subarray of size K has a given Sum // Function to check if any Subarray // of size K has a given Sum function checkSubarraySum(arr, n, k, sum) { // Check for first window let curr_sum = 0; for (let i = 0; i < k; i++) curr_sum += arr[i]; if (curr_sum == sum) return true; // Consider remaining blocks // ending with j for (let j = k; j < n; j++) { curr_sum = curr_sum + arr[j] - arr[j - k]; if (curr_sum == sum) return true; } return false; } // Driver code let arr = new Array( 1, 4, 2, 10, 2, 3, 1, 0, 20 ); let k = 4; let sum = 18; let n = arr.length; if (checkSubarraySum(arr, n, k, sum)) document.write("YES"); else document.write("NO"); // This code is contributed // by _saurabh_jaiswal </script> Time Complexity: O(n)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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