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Maximum prefix sum which is equal to suffix sum such that prefix and suffix do not overlap

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  • Difficulty Level : Easy
  • Last Updated : 30 Nov, 2022
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Given an array arr[] of N Positive integers, the task is to find the largest prefix sum which is also the suffix sum and prefix and suffix do not overlap.

Examples:

Input: N = 5, arr = [1, 3, 2, 1, 4]
Output: 4
Explanation: consider prefix [1, 3] and suffix [4] which gives maximum 
prefix sum which is also suffix sum such that prefix and suffix do not overlap.

Input: N = 5, arr = [1, 3, 1, 1, 4]
Output: 5

Approach: The problem can be solved using the two-pointer technique.

Use two pointers from both ends of array and keep maintaining sum of prefix and suffix, keep moving pointers till they overlap. 

Follow the steps to solve the problem:

  • Declare and initialize two variables i = 0 and j = N – 1.
  • Declare and initialize two variables to store prefix and suffix sum, prefix = 0, suffix = 0.
  • Declare and initialize a variable result to keep the maximum possible prefix sum, result = 0.
  • while i is less than or equal to 
    • If prefix sum is less than suffix sum add array element at the ith index to prefix sum and increment value of i.
    • Else add array element at the jth index to suffix sum and decrement value of j.
    • If both of them are equal update the result variable with prefix sum.
  • Print value of the result.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the largest prefix
// whose sum is same as the suffix sum
// such that they don't overlap
int maxPrefixSum(int N, int* arr)
{
    // Pointers pointing first and
    // last index of array
    int i = 0, j = N - 1;

    // Variables to store prefix and suffixSum
    int prefixSum = 0, suffixSum = 0;

    // Variable to store result that is
    // maximum possible prefix sum
    int result = 0;

    // While prefix and suffix
    // do not overlap
    while (i <= j) {

        // If prefix sum is less than suffix sum
        // add array element at the ith index to
        // prefix sum and increment value of i.
        if (prefixSum < suffixSum) {
            prefixSum += arr[i];
            i++;
        }

        // Else add array element at the jth
        // index to suffix sum and decrement
        // value of j
        else {
            suffixSum += arr[j];
            j--;
        }

        // If both of them are equal update
        // result variable with prefix sum.
        if (prefixSum == suffixSum)
            result = prefixSum;
    }

    return result;
}

// Driver code
int main()
{
    int arr[] = { 1, 3, 1, 1, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    cout << maxPrefixSum(N, arr);
    return 0;
}

Java

// Java code to implement the approach
import java.io.*;

class GFG {
    // Function to find the largest prefix
    // whose sum is same as the suffix sum
    // such that they don't overlap
    public static int maxPrefixSum(int N, int arr[])
    {
        // Pointers pointing first and
        // last index of array
        int i = 0, j = N - 1;

        // Variables to store prefix and suffixSum
        int prefixSum = 0, suffixSum = 0;

        // Variable to store result that is
        // maximum possible prefix sum
        int result = 0;

        // While prefix and suffix
        // do not overlap
        while (i <= j) {

            // If prefix sum is less than suffix sum
            // add array element at the ith index to
            // prefix sum and increment value of i.
            if (prefixSum < suffixSum) {
                prefixSum += arr[i];
                i++;
            }

            // Else add array element at the jth
            // index to suffix sum and decrement
            // value of j
            else {
                suffixSum += arr[j];
                j--;
            }

            // If both of them are equal update
            // result variable with prefix sum.
            if (prefixSum == suffixSum)
                result = prefixSum;
        }

        return result;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 1, 3, 1, 1, 4 };
        int N = arr.length;

        // Function Call
        System.out.print(maxPrefixSum(N, arr));
    }
}

// This code is contributed by Rohit Pradhan

Python3

# Python code for the above approach
import math

# Function to find the largest prefix
# whose sum is same as the suffix sum
# such that they don't overlap
def maxPrefixSum(N, arr):

    # Pointers pointing first and
    # last index of array
    i = 0
    j = N - 1;

    # Variables to store prefix and suffixSum
    prefixSum = 0
    suffixSum = 0;

    # Variable to store result that is
    # maximum possible prefix sum
    result = 0;

    # While prefix and suffix
    # do not overlap
    while i <= j:

        # If prefix sum is less than suffix sum
        # add array element at the ith index to
        # prefix sum and increment value of i.
        if (prefixSum < suffixSum):
            prefixSum += arr[i];
            i = i + 1;
        

        # Else add array element at the jth
        # index to suffix sum and decrement
        # value of j
        else:
            suffixSum += arr[j];
            j = j - 1;
        
        # If both of them are equal update
        # result variable with prefix sum.
        if (prefixSum == suffixSum):
            result = prefixSum;
    

    return result;

# Driver code
arr = [1, 3, 1, 1, 4];
N = len(arr);

    # Function Call
print(maxPrefixSum(N, arr));
    
# This code is contributed by Potta Lokesh

C#

// C# code to implement the approach

using System;

public class GFG{

      // Function to find the largest prefix
    // whose sum is same as the suffix sum
    // such that they don't overlap
    public static int maxPrefixSum(int N, int[] arr)
    {
        // Pointers pointing first and
        // last index of array
        int i = 0, j = N - 1;
 
        // Variables to store prefix and suffixSum
        int prefixSum = 0, suffixSum = 0;
 
        // Variable to store result that is
        // maximum possible prefix sum
        int result = 0;
 
        // While prefix and suffix
        // do not overlap
        while (i <= j) {
 
            // If prefix sum is less than suffix sum
            // add array element at the ith index to
            // prefix sum and increment value of i.
            if (prefixSum < suffixSum) {
                prefixSum += arr[i];
                i++;
            }
 
            // Else add array element at the jth
            // index to suffix sum and decrement
            // value of j
            else {
                suffixSum += arr[j];
                j--;
            }
 
            // If both of them are equal update
            // result variable with prefix sum.
            if (prefixSum == suffixSum)
                result = prefixSum;
        }
 
        return result;
    }
  
    static public void Main (){

        int[] arr = { 1, 3, 1, 1, 4 };
        int N = arr.Length;
 
        // Function Call
        Console.Write(maxPrefixSum(N, arr));
    }
}
 
// This code is contributed by lokeshmvs21.

Javascript

<script>

// JavaScript implementation of the approach

    // Function to find the largest prefix
    // whose sum is same as the suffix sum
    // such that they don't overlap
    function maxPrefixSum(N, arr)
    {
        // Pointers pointing first and
        // last index of array
        let i = 0, j = N - 1;

        // Variables to store prefix and suffixSum
        let prefixSum = 0, suffixSum = 0;

        // Variable to store result that is
        // maximum possible prefix sum
        let result = 0;

        // While prefix and suffix
        // do not overlap
        while (i <= j) {

            // If prefix sum is less than suffix sum
            // add array element at the ith index to
            // prefix sum and increment value of i.
            if (prefixSum < suffixSum) {
                prefixSum += arr[i];
                i++;
            }

            // Else add array element at the jth
            // index to suffix sum and decrement
            // value of j
            else {
                suffixSum += arr[j];
                j--;
            }

            // If both of them are equal update
            // result variable with prefix sum.
            if (prefixSum == suffixSum)
                result = prefixSum;
        }

        return result;
    }
    
// Driver code

        let arr = [1, 3, 1, 1, 4];
        let N = arr.length;

        // Function Call
        document.write(maxPrefixSum(N, arr));
      
      // This code is contributed by sanjoy_62.
</script>
Output

5

Time Complexity: O(N)
Auxiliary Space: O(1)


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