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Maximize the maximum subarray sum after removing atmost one element
  • Difficulty Level : Hard
  • Last Updated : 04 May, 2021

Given an array arr[] of N integers. The task is to first find the maximum sub-array sum and then remove at most one element from the sub-array. If there are multiple sub-arrays with the maximum sub-array sum then remove at most a single element such that the maximum sum after removal is maximized. The task is to maximize the sum obtained after removal.
Note: You have to first find the maximum sub-array sum and then remove the element from that sub-array if necessary. Also after removal, the sub-array size should at least be 1.
Examples: 
 

Input: arr[] = {1, 2, 3, -2, 3} 
Output:
The maximum sub-array sum is given by the sub-array {2, 3, -2, 3} 
Hence, we can remove -2 to further maximize the sub-array sum. 
Input: arr[] = {-1, -2} 
Output: -1 
The maximum sub-array sum is from the sub-array {-1} and no removal is required. 
 

 

Approach: Use Kadane’s algorithm to find the maximum subarray sum. Once the sum has been find, re-apply Kadane’s algorithm to find the maximum sum again with some minor changes. Use two extra variables in the loop, cnt, and mini. The variable cnt counts the number of elements in sub-array and mini stores the minimum value in all the sub-arrays which have same sum as maximum sub-array. If the minimum element thus obtained is less than 0, then only we remove an element from the subarray, or else we donot. 
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the maximum sub-array sum
int maxSubArraySum(int a[], int size)
{
 
    // Initialized
    int max_so_far = INT_MIN, max_ending_here = 0;
 
    // Traverse in the array
    for (int i = 0; i < size; i++) {
 
        // Increase the sum
        max_ending_here = max_ending_here + a[i];
 
        // If sub-array sum is more than the previous
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
 
        // If sum is negative
        if (max_ending_here < 0)
            max_ending_here = 0;
    }
    return max_so_far;
}
 
// Function that returns the maximum sub-array sum
// after removing an element from the same sub-array
int maximizeSum(int a[], int n)
{
    int cnt = 0;
    int mini = INT_MAX;
    int minSubarray = INT_MAX;
 
    // Maximum sub-array sum using Kadane's Algorithm
    int sum = maxSubArraySum(a, n);
 
    int max_so_far = INT_MIN, max_ending_here = 0;
 
    // Re-apply Kadane's with minor changes
    for (int i = 0; i < n; i++) {
 
        // Increase the sum
        max_ending_here = max_ending_here + a[i];
        cnt++;
        minSubarray = min(a[i], minSubarray);
 
        // If sub-array sum is greater than the previous
        if (sum == max_ending_here) {
 
            // If elements are 0, no removal
            if (cnt == 1)
                mini = min(mini, 0);
 
            // If elements are more, then store
            // the minimum value in the sub-array
            // obtained till now
            else
                mini = min(mini, minSubarray);
        }
 
        // If sum is negative
        if (max_ending_here < 0) {
 
            // Re-initialize everything
            max_ending_here = 0;
            cnt = 0;
            minSubarray = INT_MAX;
        }
    }
 
    return sum - mini;
}
 
// Driver code
int main()
{
    int a[] = { 1, 2, 3, -2, 3 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << maximizeSum(a, n);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
     
// Function to return the maximum sub-array sum
static int maxSubArraySum(int a[], int size)
{
 
    // Initialized
    int max_so_far = Integer.MIN_VALUE,
        max_ending_here = 0;
 
    // Traverse in the array
    for (int i = 0; i < size; i++)
    {
 
        // Increase the sum
        max_ending_here = max_ending_here + a[i];
 
        // If sub-array sum is more than the previous
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
 
        // If sum is negative
        if (max_ending_here < 0)
            max_ending_here = 0;
    }
    return max_so_far;
}
 
// Function that returns the maximum sub-array sum
// after removing an element from the same sub-array
static int maximizeSum(int a[], int n)
{
    int cnt = 0;
    int mini = Integer.MAX_VALUE;
    int minSubarray = Integer.MAX_VALUE;
 
    // Maximum sub-array sum
    // using Kadane's Algorithm
    int sum = maxSubArraySum(a, n);
 
    int max_so_far = Integer.MIN_VALUE,
        max_ending_here = 0;
 
    // Re-apply Kadane's with minor changes
    for (int i = 0; i < n; i++)
    {
 
        // Increase the sum
        max_ending_here = max_ending_here + a[i];
        cnt++;
        minSubarray = Math.min(a[i], minSubarray);
 
        // If sub-array sum is greater than the previous
        if (sum == max_ending_here)
        {
 
            // If elements are 0, no removal
            if (cnt == 1)
                mini = Math.min(mini, 0);
 
            // If elements are more, then store
            // the minimum value in the sub-array
            // obtained till now
            else
                mini = Math.min(mini, minSubarray);
        }
 
        // If sum is negative
        if (max_ending_here < 0)
        {
 
            // Re-initialize everything
            max_ending_here = 0;
            cnt = 0;
            minSubarray = Integer.MAX_VALUE;
        }
    }
 
    return sum - mini;
}
 
// Driver code
public static void main(String[] args)
{
    int a[] = { 1, 2, 3, -2, 3 };
    int n = a.length;
    System.out.println(maximizeSum(a, n));
}
}
 
// This code is contributed by Code_Mech

Python3




# Python3 implementation of the approach
import sys;
 
# Function to return the maximum sub-array sum
def maxSubArraySum(a, size) :
 
    # Initialized
    max_so_far = -(sys.maxsize - 1);
    max_ending_here = 0;
 
    # Traverse in the array
    for i in range(size) :
 
        # Increase the sum
        max_ending_here = max_ending_here + a[i];
 
        # If sub-array sum is more than the previous
        if (max_so_far < max_ending_here) :
            max_so_far = max_ending_here;
 
        # If sum is negative
        if (max_ending_here < 0) :
            max_ending_here = 0;
     
    return max_so_far;
 
# Function that returns the maximum
# sub-array sum after removing an
# element from the same sub-array
def maximizeSum(a, n) :
 
    cnt = 0;
    mini = sys.maxsize;
    minSubarray = sys.maxsize;
 
    # Maximum sub-array sum using
    # Kadane's Algorithm
    sum = maxSubArraySum(a, n);
 
    max_so_far = -(sys.maxsize - 1);
    max_ending_here = 0;
 
    # Re-apply Kadane's with minor changes
    for i in range(n) :
 
        # Increase the sum
        max_ending_here = max_ending_here + a[i];
        cnt += 1;
        minSubarray = min(a[i], minSubarray);
 
        # If sub-array sum is greater
        # than the previous
        if (sum == max_ending_here) :
 
            # If elements are 0, no removal
            if (cnt == 1) :
                mini = min(mini, 0);
 
            # If elements are more, then store
            # the minimum value in the sub-array
            # obtained till now
            else :
                mini = min(mini, minSubarray);
         
        # If sum is negative
        if (max_ending_here < 0) :
 
            # Re-initialize everything
            max_ending_here = 0;
            cnt = 0;
            minSubarray = sys.maxsize;
 
    return sum - mini;
 
# Driver code
if __name__ == "__main__" :
 
    a = [ 1, 2, 3, -2, 3 ];
    n = len(a)
     
    print(maximizeSum(a, n));
 
# This code is contributed by Ryuga

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the maximum sub-array sum
static int maxSubArraySum(int []a, int size)
{
 
    // Initialized
    int max_so_far = int.MinValue,
        max_ending_here = 0;
 
    // Traverse in the array
    for (int i = 0; i < size; i++)
    {
 
        // Increase the sum
        max_ending_here = max_ending_here + a[i];
 
        // If sub-array sum is more than the previous
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
 
        // If sum is negative
        if (max_ending_here < 0)
            max_ending_here = 0;
    }
    return max_so_far;
}
 
// Function that returns the maximum sub-array sum
// after removing an element from the same sub-array
static int maximizeSum(int []a, int n)
{
    int cnt = 0;
    int mini = int.MaxValue;
    int minSubarray = int.MaxValue;
 
    // Maximum sub-array sum
    // using Kadane's Algorithm
    int sum = maxSubArraySum(a, n);
 
    int max_so_far = int.MinValue,
        max_ending_here = 0;
 
    // Re-apply Kadane's with minor changes
    for (int i = 0; i < n; i++)
    {
 
        // Increase the sum
        max_ending_here = max_ending_here + a[i];
        cnt++;
        minSubarray = Math.Min(a[i], minSubarray);
 
        // If sub-array sum is greater than the previous
        if (sum == max_ending_here)
        {
 
            // If elements are 0, no removal
            if (cnt == 1)
                mini = Math.Min(mini, 0);
 
            // If elements are more, then store
            // the minimum value in the sub-array
            // obtained till now
            else
                mini = Math.Min(mini, minSubarray);
        }
 
        // If sum is negative
        if (max_ending_here < 0)
        {
 
            // Re-initialize everything
            max_ending_here = 0;
            cnt = 0;
            minSubarray = int.MaxValue;
        }
    }
 
    return sum - mini;
}
 
// Driver code
public static void Main(String[] args)
{
    int []a = { 1, 2, 3, -2, 3 };
    int n = a.Length;
    Console.WriteLine(maximizeSum(a, n));
}
}
 
// This code has been contributed by 29AjayKumar

PHP




<?php
// PHP implementation of the approach
 
// Function to return the maximum sub-array sum
function maxSubArraySum($a, $size)
{
 
    // Initialized
    $max_so_far = PHP_INT_MIN;
    $max_ending_here = 0;
 
    // Traverse in the array
    for ( $i = 0; $i < $size; $i++)
    {
 
        // Increase the sum
        $max_ending_here = $max_ending_here + $a[$i];
 
        // If sub-array sum is more than the previous
        if ($max_so_far < $max_ending_here)
            $max_so_far = $max_ending_here;
 
        // If sum is negative
        if ($max_ending_here < 0)
            $max_ending_here = 0;
    }
    return $max_so_far;
}
 
// Function that returns the maximum sub-array sum
// after removing an element from the same sub-array
function maximizeSum($a, $n)
{
    $cnt = 0;
    $mini = PHP_INT_MAX;
    $minSubarray = PHP_INT_MAX;
 
    // Maximum sub-array sum using Kadane's Algorithm
    $sum = maxSubArraySum($a, $n);
 
    $max_so_far = PHP_INT_MIN;
    $max_ending_here = 0;
 
    // Re-apply Kadane's with minor changes
    for ($i = 0; $i < $n; $i++)
    {
 
        // Increase the sum
        $max_ending_here = $max_ending_here + $a[$i];
        $cnt++;
        $minSubarray = min($a[$i], $minSubarray);
 
        // If sub-array sum is greater than the previous
        if ($sum == $max_ending_here)
        {
 
            // If elements are 0, no removal
            if ($cnt == 1)
                $mini = min($mini, 0);
 
            // If elements are more, then store
            // the minimum value in the sub-array
            // obtained till now
            else
                $mini = min($mini, $minSubarray);
        }
 
        // If sum is negative
        if ($max_ending_here < 0)
        {
 
            // Re-initialize everything
            $max_ending_here = 0;
            $cnt = 0;
            $minSubarray = PHP_INT_MAX;
        }
    }
    return $sum - $mini;
}
 
    // Driver code
    $a = array( 1, 2, 3, -2, 3 );
    $n = sizeof($a) / sizeof($a[0]);
    echo maximizeSum($a, $n);
 
// This code is contributed by Tushil.
?>

Javascript




<script>
// Java script implementation of the approach
 
// Function to return the maximum sub-array sum
function maxSubArraySum(a,size)
{
 
    // Initialized
    let max_so_far = Number.MIN_VALUE,
        max_ending_here = 0;
 
    // Traverse in the array
    for (let i = 0; i < size; i++)
    {
 
        // Increase the sum
        max_ending_here = max_ending_here + a[i];
 
        // If sub-array sum is more than the previous
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
 
        // If sum is negative
        if (max_ending_here < 0)
            max_ending_here = 0;
    }
    return max_so_far;
}
 
// Function that returns the maximum sub-array sum
// after removing an element from the same sub-array
function maximizeSum(a,n)
{
    let cnt = 0;
    let mini = Number.MAX_VALUE;
    let minSubarray = Number.MAX_VALUE;
 
    // Maximum sub-array sum
    // using Kadane's Algorithm
    let sum = maxSubArraySum(a, n);
 
    let max_so_far = Number.MIN_VALUE,
        max_ending_here = 0;
 
    // Re-apply Kadane's with minor changes
    for (let i = 0; i < n; i++)
    {
 
        // Increase the sum
        max_ending_here = max_ending_here + a[i];
        cnt++;
        minSubarray = Math.min(a[i], minSubarray);
 
        // If sub-array sum is greater than the previous
        if (sum == max_ending_here)
        {
 
            // If elements are 0, no removal
            if (cnt == 1)
                mini = Math.min(mini, 0);
 
            // If elements are more, then store
            // the minimum value in the sub-array
            // obtained till now
            else
                mini = Math.min(mini, minSubarray);
        }
 
        // If sum is negative
        if (max_ending_here < 0)
        {
 
            // Re-initialize everything
            max_ending_here = 0;
            cnt = 0;
            minSubarray = Integer.MAX_VALUE;
        }
    }
 
    return sum - mini;
}
 
// Driver code
 
    let a = [ 1, 2, 3, -2, 3 ];
    let n = a.length;
    document.write(maximizeSum(a, n));
 
 
// This code is contributed by sravan kumar Gottumukkala
</script>
Output: 
9

 

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