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:

C++

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// 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;
}

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Java

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// Java program for Split the array into three 
// subarrays such that summation of first 
// and third subarray is equal and maximum 
import java.util.HashMap;
import java.util.Map;
  
class GfG
{
  
    // Fucntion to return the sum 
    // of the first subarray 
    static int sumFirst(int a[], int n) 
    
        HashMap<Integer, Integer> mp = new HashMap<>(); 
        int suf = 0
      
        // calculate the suffix sum 
        for (int i = n - 1; i >= 0; i--) 
        
            suf += a[i]; 
            mp.put(suf, i); 
        
      
        int pre = 0, 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.containsKey(pre) && mp.get(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
    public static void main(String []args)
    {
          
        int a[] = { 1, 3, 2, 1, 4 }; 
        int n = a.length; 
        System.out.println(sumFirst(a, n));
    }
}
  
// This code is contributed by Rituraj Jain

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Python3

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# Python 3 program for Split the array into three
# subarrays such that summation of first
# and third subarray is equal and maximum
  
# Fucntion to return the sum of
# the first subarray
def sumFirst(a, n):
    mp = {i:0 for i in range(7)}
    suf = 0
    i = n - 1
      
    # calculate the suffix sum
    while(i >= 0):
        suf += a[i]
        mp[suf] = i
        i -= 1
  
    pre = 0
    maxi = -1
  
    # iterate from beginning
    for i in range(n):
          
        # 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
if __name__ == '__main__':
    a = [1, 3, 2, 1, 4]
    n = len(a)
    print(sumFirst(a, n))
      
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# program for Split the array into three 
// subarrays such that summation of first 
// and third subarray is equal and maximum 
using System;
using System.Collections.Generic;
      
class GfG
{
  
    // Fucntion to return the sum 
    // of the first subarray 
    static int sumFirst(int []a, int n) 
    
        Dictionary<int,int> mp = new Dictionary<int,int>();
        int suf = 0; 
      
        // calculate the suffix sum 
        for (int i = n - 1; i >= 0; i--) 
        
            suf += a[i]; 
            mp.Add(suf, i);
            if(mp.ContainsKey(suf))
            {
                mp.Remove(suf);
            }
                mp.Add(suf, i);
        
      
        int pre = 0, 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.ContainsKey(pre) && 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
    public static void Main(String []args)
    {
          
        int []a = { 1, 3, 2, 1, 4 }; 
        int n = a.Length; 
        Console.WriteLine(sumFirst(a, n));
    }
}
  
// This code is contributed by Rajput-Ji

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Output:

4


My Personal Notes arrow_drop_up

Striver(underscore)79 at Codechef and codeforces D

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