Maximize count of non-overlapping subarrays with sum K

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:



  1. Initialize a set to store all the prefix sums obtained up to the current element.
  2. 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.
  3. 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.
  4. Repeat the above steps until the entire array is traversed. Finally, print the value of res.

C++14

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// 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);
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output: 

3

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

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