Split array to three subarrays such that sum of first and third subarray is equal and maximum

Given an array of N integers, the task is to print the sum of the first subarray by splitting the array into exactly three subarrays such that the sum of the first and third subarray elements are equal and the maximum.

Note: All the elements must belong to a subarray and the subarrays can also be empty.

Examples:

Input: a[] = {1, 3, 1, 1, 4}
Output: 5
Split the N numbers to [1, 3, 1], [] and [1, 4]

Input: a[] = {1, 3, 2, 1, 4}
Output: 4
Split the N numbers to [1, 3], [2, 1] and [4]



A naive approach is to check for all possible partitions and use the prefix-sum concept to find out the partitions. The partition which gives the maximum sum of the first subarray will be the answer.

An efficient approach is as follows:

  • Store the prefix sum and suffix sum of the N numbers.
  • Hash the suffix sum’s index using a unordered_map in C++ or Hash-map in Java.
  • Iterate from the beginning of the array, and check if the prefix sum exists in the suffix array beyond the current index i.
  • If it does, then check for the previous maximum value and update accordingly.

Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for Split the array into three
// subarrays such that summation of first
// and third subarray is equal and maximum
#include <bits/stdc++.h>
using namespace std;
  
// Fucntion to return the sum of
// the first subarray
int sumFirst(int a[], int n)
{
    unordered_map<int, int> mp;
    int suf = 0;
  
    // calculate the suffix sum
    for (int i = n - 1; i >= 0; i--) {
        suf += a[i];
        mp[suf] = i;
    }
  
    int pre = 0;
    int maxi = -1;
  
    // iterate from beginning
    for (int i = 0; i < n; i++) {
  
        // prefix sum
        pre += a[i];
  
        // check if it exists beyond i
        if (mp[pre] > i) {
  
            // if greater then previous
            // then update maximum
            if (pre > maxi) {
                maxi = pre;
            }
        }
    }
  
    // First and second subarray empty
    if (maxi == -1)
        return 0;
  
    // partition done
    else
        return maxi;
}
  
// Driver Code
int main()
{
    int a[] = { 1, 3, 2, 1, 4 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << sumFirst(a, n);
    return 0;
}

chevron_right


Output:

4


My Personal Notes arrow_drop_up

Striver(underscore)79 at Codechef and codeforces D

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.




Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.