Maximum sum of array after removing a positive or negative subarray

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Given an array arr[] of N non-zero integers, the task is to find the maximum sum of the array by removing exactly one contiguous set of positive or negative elements.

Examples:

Input: arr[] = {-2, -3, 4, -1, -2, 1, 5, -3}
Output: 4
Explanation: Maximum array sum can be obtained by removing subarray arr[0, 1] since arr[0, 1] has same type of elements i.e., negative. Thus, the required sum is 4.

Input: arr[] = {2, -10, 4, 2, -8, -7}
Output: -2
Explanation: Maximum array sum can be obtained by removing subarray arr[4, 5] since arr[4, 5] has same type of elements i.e., negative. Thus, the required sum is -2.

Approach: The given problem can be solved based on the following observation i.e to obtain the maximum sum, a contiguous set of negative elements are to be removed since removing positive elements will reduce the array sum. However, if there are no negative elements then remove the smallest element of the array. Follow the steps to solve the problem:

Traverse the array, arr[] and store the total sum of the array in the variable, say sum.
Store the maximum contiguous negative sum in a variable, say max_neg.
If there are no negative elements in the array, then update max_neg to the smallest element of an array.
Update the value of sum to (sum – max_neg).
Print the value of sum as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum sum of
// array after removing either the contiguous
// positive or negative elements
void maxSum(int arr[], int n)
{
    // Store the total sum of array
    int sum = 0;
 
    // Store the maximum contiguous
    // negative sum
    int max_neg = INT_MAX;
 
    // Store the sum of current
    // contiguous negative elements
    int tempsum = 0;
 
    // Store the minimum element of array
    int small = INT_MAX;
 
    // Traverse the array, arr[]
    for (int i = 0; i < n; i++) {
 
        // Update the overall sum
        sum += arr[i];
 
        // Store minimum element of array
        small = min(small, arr[i]);
 
        // If arr[i] is positive
        if (arr[i] > 0) {
 
            // Update temp_sum to 0
            tempsum = 0;
        }
 
        else {
 
            // Add arr[i] to temp_sum
            tempsum += arr[i];
        }
 
        // Update max_neg
        max_neg = min(max_neg, tempsum);
    }
 
    // If no negative element in array
    // then remove smallest positive element
    if (max_neg == 0) {
        max_neg = small;
    }
 
    // Print the required sum
    cout << sum - max_neg;
}
 
// Driver Code
int main()
{
    // Given Input
    int arr[] = { -2, -3, 4, -1, -2, 1, 5, -3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    maxSum(arr, n);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to find the maximum sum of
// array after removing either the contiguous
// positive or negative elements
static void maxSum(int arr[], int n)
{
     
    // Store the total sum of array
    int sum = 0;
 
    // Store the maximum contiguous
    // negative sum
    int max_neg = Integer.MAX_VALUE;
 
    // Store the sum of current
    // contiguous negative elements
    int tempsum = 0;
 
    // Store the minimum element of array
    int small = Integer.MAX_VALUE;
 
    // Traverse the array, arr[]
    for(int i = 0; i < n; i++) 
    {
         
        // Update the overall sum
        sum += arr[i];
 
        // Store minimum element of array
        small = Math.min(small, arr[i]);
 
        // If arr[i] is positive
        if (arr[i] > 0) 
        {
             
            // Update temp_sum to 0
            tempsum = 0;
        }
        else
        {
             
            // Add arr[i] to temp_sum
            tempsum += arr[i];
        }
 
        // Update max_neg
        max_neg = Math.min(max_neg, tempsum);
    }
 
    // If no negative element in array
    // then remove smallest positive element
    if (max_neg == 0) 
    {
        max_neg = small;
    }
 
    // Print the required sum
    System.out.println(sum - max_neg);
}
 
// Driver Code
public static void main(String[] args) 
{
     
    // Given Input
    int arr[] = { -2, -3, 4, -1, -2, 1, 5, -3 };
    int n = arr.length;
 
    // Function Call
    maxSum(arr, n);
}
}
 
// This code is contributed by Dharanendra L V.

Python3

# python 3 program for the above approach
 
import sys
# Function to find the maximum sum of
# array after removing either the contiguous
# positive or negative elements
def maxSum(arr, n):
   
    # Store the total sum of array
    sum = 0
 
    # Store the maximum contiguous
    # negative sum
    max_neg = sys.maxsize
 
    # Store the sum of current
    # contiguous negative elements
    tempsum = 0
 
    # Store the minimum element of array
    small = sys.maxsize
 
    # Traverse the array, arr[]
    for i in range(n):
        # Update the overall sum
        sum += arr[i]
 
        # Store minimum element of array
        small = min(small, arr[i])
 
        # If arr[i] is positive
        if (arr[i] > 0):
            # Update temp_sum to 0
            tempsum = 0
 
        else:
 
            # Add arr[i] to temp_sum
            tempsum += arr[i]
 
        # Update max_neg
        max_neg = min(max_neg, tempsum)
 
    # If no negative element in array
    # then remove smallest positive element
    if (max_neg == 0):
        max_neg = small
 
    # Print the required sum
    print(sum - max_neg)
 
# Driver Code
if __name__ == '__main__':
   
    # Given Input
    arr = [-2, -3, 4, -1, -2, 1, 5, -3]
    n = len(arr)
 
    # Function Call
    maxSum(arr, n)
     
    # This code is contributed by bgangwar59.

Javascript

<script>
// Javascript program for the above approach
 
 
// Function to find the maximum sum of
// array after removing either the contiguous
// positive or negative elements
function maxSum(arr, n) {
    // Store the total sum of array
    let sum = 0;
 
    // Store the maximum contiguous
    // negative sum
    let max_neg = Number.MAX_SAFE_INTEGER;
 
    // Store the sum of current
    // contiguous negative elements
    let tempsum = 0;
 
    // Store the minimum element of array
    let small = Number.MAX_SAFE_INTEGER;
 
    // Traverse the array, arr[]
    for (let i = 0; i < n; i++) {
 
        // Update the overall sum
        sum += arr[i];
 
        // Store minimum element of array
        small = Math.min(small, arr[i]);
 
        // If arr[i] is positive
        if (arr[i] > 0) {
 
            // Update temp_sum to 0
            tempsum = 0;
        }
 
        else {
 
            // Add arr[i] to temp_sum
            tempsum += arr[i];
        }
 
        // Update max_neg
        max_neg = Math.min(max_neg, tempsum);
    }
 
    // If no negative element in array
    // then remove smallest positive element
    if (max_neg == 0) {
        max_neg = small;
    }
 
    // Print the required sum
    document.write(sum - max_neg);
}
 
// Driver Code
 
// Given Input
let arr = [-2, -3, 4, -1, -2, 1, 5, -3];
let n = arr.length;
 
// Function Call
maxSum(arr, n);
 
// This code is contributed by gfgking.
</script>

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the maximum sum of
// array after removing either the contiguous
// positive or negative elements
static void maxSum(int []arr, int n)
{
     
    // Store the total sum of array
    int sum = 0;
 
    // Store the maximum contiguous
    // negative sum
    int max_neg = Int32.MaxValue;
 
    // Store the sum of current
    // contiguous negative elements
    int tempsum = 0;
 
    // Store the minimum element of array
    int small = Int32.MaxValue;
 
    // Traverse the array, arr[]
    for(int i = 0; i < n; i++) 
    {
         
        // Update the overall sum
        sum += arr[i];
 
        // Store minimum element of array
        small = Math.Min(small, arr[i]);
 
        // If arr[i] is positive
        if (arr[i] > 0) 
        {
             
            // Update temp_sum to 0
            tempsum = 0;
        }
        else
        {
             
            // Add arr[i] to temp_sum
            tempsum += arr[i];
        }
 
        // Update max_neg
        max_neg = Math.Min(max_neg, tempsum);
    }
 
    // If no negative element in array
    // then remove smallest positive element
    if (max_neg == 0) 
    {
        max_neg = small;
    }
 
    // Print the required sum
    Console.Write(sum - max_neg);
}
 
// Driver Code
public static void Main(String[] args) 
{
     
    // Given Input
    int []arr = { -2, -3, 4, -1, -2, 1, 5, -3 };
    int n = arr.Length;
 
    // Function Call
    maxSum(arr, n);
}
}
 
// This code is contributed by shivanisinghss2110

Output

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

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