Maximize count of non-overlapping subarrays with sum K

Last Updated : 11 Jul, 2022

Given an array arr[] and an integer K, the task is to print the maximum number of non-overlapping subarrays with a sum equal to K.

Examples:

Input: arr[] = {-2, 6, 6, 3, 5, 4, 1, 2, 8}, K = 10
Output: 3
Explanation: All possible non-overlapping subarrays with sum K(= 10) are {-2, 6, 6}, {5, 4, 1}, {2, 8}. Therefore, the required count is 3.

Input: arr[] = {1, 1, 1}, K = 2
Output: 1

Approach: The problem can be solved using the concept of prefix sum. Follow the below steps to solve the problem:

Initialize a set to store all the prefix sums obtained up to the current element.
Initialize variables prefixSum and res, to store the prefix sum of the current subarray and the count of subarrays with a sum equal to K respectively.
Iterate over the array and for each array element, update prefixSum by adding to it the current element. Now, check if the value prefixSum – K is already present in the set or not. If found to be true, increment res, clear the set, and reset the value of prefixSum.
Repeat the above steps until the entire array is traversed. Finally, print the value of res.

C++14

   // C++ Program to implement 
// the above approach 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to count the maximum 
// number of subarrays with sum K 
int CtSubarr(int arr[], int N, int K) 
{ 
  
    // Stores all the distinct 
    // prefixSums obtained 
    unordered_set<int> st; 
  
    // Stores the prefix sum 
    // of the current subarray 
    int prefixSum = 0; 
  
    st.insert(prefixSum); 
  
    // Stores the count of 
    // subarrays with sum K 
    int res = 0; 
  
    for (int i = 0; i < N; i++) { 
        prefixSum += arr[i]; 
  
        // If a subarray with sum K 
        // is already found 
        if (st.count(prefixSum - K)) { 
  
            // Increase count 
            res += 1; 
  
            // Reset prefix sum 
            prefixSum = 0; 
  
            // Clear the set 
            st.clear(); 
            st.insert(0); 
        } 
  
        // Insert the prefix sum 
        st.insert(prefixSum); 
    } 
    return res; 
} 
  
// Driver Code 
int main() 
{ 
    int arr[] = { -2, 6, 6, 3, 5, 4, 1, 2, 8 }; 
    int N = sizeof(arr) / sizeof(arr[0]); 
    int K = 10; 
    cout << CtSubarr(arr, N, K); 
}
 
  

Java

   // Java Program to implement 
// the above approach 
import java.util.*; 
class GFG{ 
    // Function to count the maximum 
    // number of subarrays with sum K 
    static int CtSubarr(int[] arr,  
                        int N, int K) 
    { 
        // Stores all the distinct 
        // prefixSums obtained 
        Set<Integer> st = new HashSet<Integer>(); 
  
        // Stores the prefix sum 
        // of the current subarray 
        int prefixSum = 0; 
  
        st.add(prefixSum); 
  
        // Stores the count of 
        // subarrays with sum K 
        int res = 0; 
  
        for (int i = 0; i < N; i++)  
        { 
            prefixSum += arr[i]; 
  
            // If a subarray with sum K 
            // is already found 
            if (st.contains(prefixSum - K))  
            { 
                // Increase count 
                res += 1; 
  
                // Reset prefix sum 
                prefixSum = 0; 
  
                // Clear the set 
                st.clear(); 
                st.add(0); 
            } 
  
            // Insert the prefix sum 
            st.add(prefixSum); 
        } 
        return res; 
    } 
  
    // Driver Code 
    public static void main(String[] args) 
    { 
        int arr[] = {-2, 6, 6, 3,  
                     5, 4, 1, 2, 8}; 
        int N = arr.length; 
        int K = 10; 
        System.out.println(CtSubarr(arr, N, K)); 
    } 
} 
  
// This code is contributed by Chitranayal
 
  

Python3

   # Python3 program to implement 
# the above approach 
  
# Function to count the maximum  
# number of subarrays with sum K 
def CtSubarr(arr, N, K): 
  
    # Stores all the distinct 
    # prefixSums obtained 
    st = set() 
  
    # Stores the prefix sum 
    # of the current subarray 
    prefixSum = 0
  
    st.add(prefixSum) 
  
    # Stores the count of 
    # subarrays with sum K 
    res = 0
  
    for i in range(N): 
        prefixSum += arr[i] 
  
        # If a subarray with sum K 
        # is already found 
        if((prefixSum - K) in st): 
  
            # Increase count 
            res += 1
  
            # Reset prefix sum 
            prefixSum = 0
  
            # Clear the set 
            st.clear() 
            st.add(0) 
  
        # Insert the prefix sum 
        st.add(prefixSum) 
  
    return res 
  
# Driver Code 
arr = [ -2, 6, 6, 3, 5, 4, 1, 2, 8 ] 
N = len(arr) 
K = 10
  
# Function call 
print(CtSubarr(arr, N, K)) 
  
# This code is contributed by Shivam Singh 
 
  

C#

   // C# program to implement 
// the above approach 
using System; 
using System.Collections.Generic; 
  
class GFG{ 
      
// Function to count the maximum 
// number of subarrays with sum K 
static int CtSubarr(int[] arr,  
                    int N, int K) 
{ 
      
    // Stores all the distinct 
    // prefixSums obtained 
    HashSet<int> st = new HashSet<int>(); 
  
    // Stores the prefix sum 
    // of the current subarray 
    int prefixSum = 0; 
  
    st.Add(prefixSum); 
  
    // Stores the count of 
    // subarrays with sum K 
    int res = 0; 
  
    for(int i = 0; i < N; i++)  
    { 
        prefixSum += arr[i]; 
  
        // If a subarray with sum K 
        // is already found 
        if (st.Contains(prefixSum - K))  
        { 
              
            // Increase count 
            res += 1; 
  
            // Reset prefix sum 
            prefixSum = 0; 
  
            // Clear the set 
            st.Clear(); 
            st.Add(0); 
        } 
  
        // Insert the prefix sum 
        st.Add(prefixSum); 
    } 
    return res; 
} 
  
// Driver Code 
public static void Main(String[] args) 
{ 
    int []arr = { -2, 6, 6, 3,  
                   5, 4, 1, 2, 8}; 
    int N = arr.Length; 
    int K = 10; 
      
    Console.WriteLine(CtSubarr(arr, N, K)); 
} 
} 
  
// This code is contributed by 29AjayKumar
 
  

Javascript

   <script> 
  
// Javascript Program to implement 
// the above approach 
  
// Function to count the maximum 
// number of subarrays with sum K 
function CtSubarr(arr, N, K) 
{ 
  
    // Stores all the distinct 
    // prefixSums obtained 
    var st = new Set(); 
  
    // Stores the prefix sum 
    // of the current subarray 
    var prefixSum = 0; 
  
    st.add(prefixSum); 
  
    // Stores the count of 
    // subarrays with sum K 
    var res = 0; 
  
    for (var i = 0; i < N; i++) { 
        prefixSum += arr[i]; 
  
        // If a subarray with sum K 
        // is already found 
        if (st.has(prefixSum - K)) { 
  
            // Increase count 
            res += 1; 
  
            // Reset prefix sum 
            prefixSum = 0; 
  
            // Clear the set 
            st = new Set(); 
            st.add(0); 
        } 
  
        // Insert the prefix sum 
        st.add(prefixSum); 
    } 
    return res; 
} 
  
// Driver Code 
var arr = [-2, 6, 6, 3, 5, 4, 1, 2, 8]; 
var N = arr.length; 
var K = 10; 
document.write( CtSubarr(arr, N, K)); 
  
// This code is contributed by importantly. 
</script>
 
  

Output:

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

Count maximum non-overlapping subarrays with given sum

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Maximize count of non-overlapping subarrays with sum K

C++14

Java

Python3

C#

Javascript

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