Count maximum non-overlapping subarrays with given sum

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Given an array arr[] consisting of N integers and an integer target, the task is to find the maximum number of non-empty non-overlapping subarrays such that the sum of array elements in each subarray is equal to the target.

Examples:

Input: arr[] = {2, -1, 4, 3, 6, 4, 5, 1}, target = 6
Output: 3
Explanation:
Subarrays {-1, 4, 3}, {6} and {5, 1} have sum equal to target(= 6).

Input: arr[] = {2, 2, 2, 2, 2}, target = 4
Output: 2

Approach: To obtain the smallest non-overlapping subarrays with the sum target, the target is to use the Prefix Sum technique. Follow the steps below to solve the problem:

Store all the sums calculated so far in a Map mp with key as the sum of the prefix till that index and value as the ending index of the subarray with that sum.
If the prefix-sum till index i, say sum, is equal to target, check if sum – target exists in the Map or not.
If sum – target exists in Map and mp[sum – target] = idx, it means that the subarray from [idx + 1, i] has sum equal to target.
Now for non-overlapping subarrays, maintain an additional variable availIdx(initially set to -1), and take the subarray from [idx + 1, i] only when mp[sum – target] ? availIdx.
Whenever such a subarray is found, increment the answer and change the value of availIdx to the current index.
Also, for non-overlapping subarrays, it is always beneficial to greedily take subarrays as small as possible. So, for every prefix-sum found, update its index in the Map, even if it already exists.
Print the value of count after completing the above steps.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count maximum number
// of non-overlapping subarrays with
// sum equals to the target
int maximumSubarrays(int arr[], int N,
                     int target)
{
    // Stores the final count
    int ans = 0;
 
    // Next subarray should start
    // from index >= availIdx
    int availIdx = -1;
 
    // Tracks the prefix sum
    int cur_sum = 0;
 
    // Map to store the prefix sum
    // for respective indices
    unordered_map<int, int> mp;
    mp[0] = -1;
 
    for (int i = 0; i < N; i++) {
 
        cur_sum += arr[i];
 
        // Check if cur_sum - target is
        // present in the array or not
        if (mp.find(cur_sum - target)
                != mp.end()
            && mp[cur_sum - target]
                   >= availIdx) {
 
            ans++;
            availIdx = i;
        }
 
        // Update the index of
        // current prefix sum
        mp[cur_sum] = i;
    }
 
    // Return the count of subarrays
    return ans;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 2, -1, 4, 3,
                  6, 4, 5, 1 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Given sum target
    int target = 6;
 
    // Function Call
    cout << maximumSubarrays(arr, N,
                             target);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to count maximum number
// of non-overlapping subarrays with
// sum equals to the target
static int maximumSubarrays(int arr[], int N,
                            int target)
{
     
    // Stores the final count
    int ans = 0;
 
    // Next subarray should start
    // from index >= availIdx
    int availIdx = -1;
 
    // Tracks the prefix sum
    int cur_sum = 0;
 
    // Map to store the prefix sum
    // for respective indices
    HashMap<Integer,
            Integer> mp = new HashMap<Integer,
                                      Integer>();
    mp.put(0, 1);
 
    for(int i = 0; i < N; i++)
    {
        cur_sum += arr[i];
 
        // Check if cur_sum - target is
        // present in the array or not
        if (mp.containsKey(cur_sum - target) && 
            mp.get(cur_sum - target) >= availIdx)
        {
            ans++;
            availIdx = i;
        }
         
        // Update the index of
        // current prefix sum
        mp.put(cur_sum, i);
    }
 
    // Return the count of subarrays
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = { 2, -1, 4, 3,
                  6, 4, 5, 1 };
 
    int N = arr.length;
 
    // Given sum target
    int target = 6;
 
    // Function call
    System.out.print(maximumSubarrays(arr, N, 
                                      target));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
 
# Function to count maximum number
# of non-overlapping subarrays with
# sum equals to the target
def maximumSubarrays(arr, N, target):
     
    # Stores the final count
    ans = 0
 
    # Next subarray should start
    # from index >= availIdx
    availIdx = -1
 
    # Tracks the prefix sum
    cur_sum = 0
 
    # Map to store the prefix sum
    # for respective indices
    mp = {}
    mp[0] = -1
 
    for i in range(N):
        cur_sum += arr[i]
 
        # Check if cur_sum - target is
        # present in the array or not
        if ((cur_sum - target) in mp and
          mp[cur_sum - target] >= availIdx):
            ans += 1
            availIdx = i
 
        # Update the index of
        # current prefix sum
        mp[cur_sum] = i
 
    # Return the count of subarrays
    return ans
 
# Driver Code
if __name__ == '__main__':
     
    # Given array arr[]
    arr = [ 2, -1, 4, 3,
            6, 4, 5, 1 ]
 
    N = len(arr)
 
    # Given sum target
    target = 6
 
    # Function call
    print(maximumSubarrays(arr, N, target))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Function to count maximum number
// of non-overlapping subarrays with
// sum equals to the target
static int maximumSubarrays(int []arr, int N,
                            int target)
{
  // Stores the readonly count
  int ans = 0;
 
  // Next subarray should start
  // from index >= availIdx
  int availIdx = -1;
 
  // Tracks the prefix sum
  int cur_sum = 0;
 
  // Map to store the prefix sum
  // for respective indices
  Dictionary<int,
             int> mp = new Dictionary<int,
                                      int>();
  mp.Add(0, 1);
 
  for(int i = 0; i < N; i++)
  {
    cur_sum += arr[i];
 
    // Check if cur_sum - target is
    // present in the array or not
    if (mp.ContainsKey(cur_sum - target) && 
        mp[cur_sum - target] >= availIdx)
    {
      ans++;
      availIdx = i;
    }
 
    // Update the index of
    // current prefix sum
    if(mp.ContainsKey(cur_sum))
      mp[cur_sum] = i;
    else
      mp.Add(cur_sum, i);
  }
 
  // Return the count of subarrays
  return ans;
}
 
// Driver Code
public static void Main(String[] args)
{    
  // Given array []arr
  int []arr = {2, -1, 4, 3,
               6, 4, 5, 1};
   
  int N = arr.Length;
 
  // Given sum target
  int target = 6;
 
  // Function call
  Console.Write(maximumSubarrays(arr, N, 
                                 target));
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>
 
// JavaScript program for the above approach
 
// Function to count maximum number
// of non-overlapping subarrays with
// sum equals to the target
function maximumSubarrays(arr, N, target)
{
    // Stores the final count
    var ans = 0;
 
    // Next subarray should start
    // from index >= availIdx
    var availIdx = -1;
 
    // Tracks the prefix sum
    var cur_sum = 0;
 
    // Map to store the prefix sum
    // for respective indices
    var mp = new Map();
    mp.set(0, 1);
 
    for (var i = 0; i < N; i++) {
 
        cur_sum += arr[i];
 
        // Check if cur_sum - target is
        // present in the array or not
        if (mp.has(cur_sum - target)
            && mp.get(cur_sum - target)
                   >= availIdx) {
 
            ans++;
            availIdx = i;
        }
 
        // Update the index of
        // current prefix sum
        mp.set(cur_sum , i);
    }
 
    // Return the count of subarrays
    return ans;
}
 
// Driver Code
 
// Given array arr[]
var arr = [2, -1, 4, 3,
              6, 4, 5, 1];
var N = arr.length;
 
// Given sum target
var target = 6;
 
// Function Call
document.write( maximumSubarrays(arr, N,
                         target));
 
</script>

Output:

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

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