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++ 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 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 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# 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  

<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>  

3

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

Related Topic: Subarrays, Subsequences, and Subsets in Array