Count maximum nonoverlapping subarrays with given sum (original) (raw)

Count maximum non-overlapping subarrays with given sum

Last Updated : 24 May, 2021

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:

  1. 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.
  2. If the prefix-sum till index i, say sum, is equal to target, check if sum - target exists in the Map or not.
  3. 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.
  4. 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.
  5. Whenever such a subarray is found, increment the answer and change the value of availIdx to the current index.
  6. 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.
  7. 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

`

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

Similar Reads