Maximize array sum by alternating the signs of adjacent elements

Given an array, arr[] of size N, the task is to find the maximum possible sum of array elements by alternating the signs of adjacent array elements.

Examples:

Input: arr[] = { -2, 1, 0 } 
Output: 3 
Explanation:
Alternating the signs of (arr[0], arr[1]) modifies arr[] to {2, -1, 0}. 
Alternating the signs of (arr[1], arr[2]) modifies arr[] to {2, 1, 0}. 
Therefore, the required output = (2 + 1 + 0) = 3, which is the maximum sum possible.

Input: arr[] = { 1, 1, -2, -4, 5 } 
Output: 13 
Explanation:
Alternating the signs of (arr[2], arr[3]) modifies arr[] to { 1, 1, 2, 4, 5 } 
Therefore, the required output = (1 + 1 + 2 + 4 + 5) = 13, which is the maximum sum possible.

Approach: The problem can be solved using Greedy technique. The idea is based on the fact that the maximum count of negative elements in the array after alternating the signs of adjacent elements can’t be greater than 1. Follow the steps below to solve the problem:

• Initialize a variable, say MaxAltSum, to store the maximum possible sum of array elements by alternating the signs of adjacent elements.
• Traverse the array and count the number of negative elements in the array.
• If count of negative elements in the array is even, then maximum possible sum possible by alternating the signs of adjacent array elements is equal to the sum of absolute value of array elements, i.e. MaxAltSum = Σabs(arr[i])
• Otherwise, maximum possible sum obtained from the array by alternating the signs of adjacent array elements equal to the sum of the absolute value of all possible array elements, except the smallest absolute value of array elements. i.e, MaxAltSum = ((Σabs(arr[i])) – 2 * X), where X is the smallest absolute value of array elements.
• Finally, print the value MaxAltSum.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach

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

// Function to find the maximum sum by alternating
// the signs of adjacent elements of the array
int findMaxSumByAlternatingSign(int arr[], int N)
{
    // Stores count of negative
    // elements in the array
    int cntNeg = 0;

    // Stores maximum sum by alternating
    // the signs of adjacent elements
    int MaxAltSum = 0;

    // Stores smallest absolute
    // value of array elements
    int SmValue = 0;

    // Stores sum of absolute
    // value of array elements
    int sum = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        // If arr[i] is
        // a negative number
        if (arr[i] < 0) {

            // Update cntNeg
            cntNeg += 1;
        }

        // Update sum
        sum += abs(arr[i]);

        // Update SmValue
        SmValue = min(SmValue,
                    abs(arr[i]));
    }

    // Update MaxAltSum
    MaxAltSum = sum;

    // If cntNeg is
    // an odd number
    if (cntNeg & 1) {

        // Update MaxAltSum
        MaxAltSum -= 2 * SmValue;
    }
    return MaxAltSum;
}

// Drivers Code
int main()
{

    int arr[] = { 1, 1, -2, -4, 5 };
    int N = sizeof(arr)
            / sizeof(arr[0]);

    cout << findMaxSumByAlternatingSign(
        arr, N);
}

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG{

// Function to find the maximum sum by alternating
// the signs of adjacent elements of the array
static int findMaxSumByAlternatingSign(int arr[],
                                       int N)
{

    // Stores count of negative
    // elements in the array
    int cntNeg = 0;

    // Stores maximum sum by alternating
    // the signs of adjacent elements
    int MaxAltSum = 0;

    // Stores smallest absolute
    // value of array elements
    int SmValue = 0;

    // Stores sum of absolute
    // value of array elements
    int sum = 0;

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // If arr[i] is
        // a negative number
        if (arr[i] < 0)
        {

            // Update cntNeg
            cntNeg += 1;
        }

        // Update sum
        sum += Math.abs(arr[i]);

        // Update SmValue
        SmValue = Math.min(SmValue,
                  Math.abs(arr[i]));
    }

    // Update MaxAltSum
    MaxAltSum = sum;

    // If cntNeg is
    // an odd number
    if (cntNeg % 2 == 1)
    {

        // Update MaxAltSum
        MaxAltSum -= 2 * SmValue;
    }
    return MaxAltSum;
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 1, -2, -4, 5 };
    int N = arr.length;

    System.out.print(findMaxSumByAlternatingSign(
    arr, N));
}
}

// This code is contributed by jana_sayantan

Python3

# Python3 program to implement
# the above approach

# Function to find the maximum sum by
# alternating the signs of adjacent
# elements of the array
def findMaxSumByAlternatingSign(arr, N):

    # Stores count of negative
    # elements in the array
    cntNeg = 0

    # Stores maximum sum by alternating
    # the signs of adjacent elements
    MaxAltSum = 0

    # Stores smallest absolute
    # value of array elements
    SmValue = 0

    # Stores sum of absolute
    # value of array elements
    sum = 0

    # Traverse the array
    for i in range(N):

        # If arr[i] is
        # a negative number
        if (arr[i] < 0):

            # Update cntNeg
            cntNeg += 1

        # Update sum
        sum += abs(arr[i])

        # Update SmValue
        SmValue = min(SmValue, abs(arr[i]))

    # Update MaxAltSum
    MaxAltSum = sum

    # If cntNeg is
    # an odd number
    if (cntNeg & 1):

        # Update MaxAltSum
        MaxAltSum -= 2 * SmValue

    return MaxAltSum

# Driver Code
if __name__ == '__main__':

    arr = [ 1, 1, -2, -4, 5 ]
    N = len(arr)

    print(findMaxSumByAlternatingSign(arr, N))

# This code is contributed by SURENDRA_GANGWAR

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to find the maximum sum by alternating
// the signs of adjacent elements of the array
static int findMaxSumByAlternatingSign(int []arr,
                                    int N)
{

    // Stores count of negative
    // elements in the array
    int cntNeg = 0;

    // Stores maximum sum by alternating
    // the signs of adjacent elements
    int MaxAltSum = 0;

    // Stores smallest absolute
    // value of array elements
    int SmValue = 0;

    // Stores sum of absolute
    // value of array elements
    int sum = 0;

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // If arr[i] is
        // a negative number
        if (arr[i] < 0)
        {

            // Update cntNeg
            cntNeg += 1;
        }

        // Update sum
        sum += Math.Abs(arr[i]);

        // Update SmValue
        SmValue = Math.Min(SmValue,
                Math.Abs(arr[i]));
    }

    // Update MaxAltSum
    MaxAltSum = sum;

    // If cntNeg is
    // an odd number
    if (cntNeg % 2 == 1)
    {

        // Update MaxAltSum
        MaxAltSum -= 2 * SmValue;
    }
    return MaxAltSum;
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 1, -2, -4, 5 };
    int N = arr.Length;

    Console.Write(findMaxSumByAlternatingSign(
    arr, N));
}
}

// This code is contributed by shivanisinghss2110

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to find the maximum sum by alternating
// the signs of adjacent elements of the array
function findMaxSumByAlternatingSign(arr, N)
{

    // Stores count of negative
    // elements in the array
    let cntNeg = 0;

    // Stores maximum sum by alternating
    // the signs of adjacent elements
    let MaxAltSum = 0;

    // Stores smallest absolute
    // value of array elements
    let SmValue = 0;

    // Stores sum of absolute
    // value of array elements
    let sum = 0;

    // Traverse the array
    for(let i = 0; i < N; i++)
    {

        // If arr[i] is
        // a negative number
        if (arr[i] < 0)
        {

            // Update cntNeg
            cntNeg += 1;
        }

        // Update sum
        sum += Math.abs(arr[i]);

        // Update SmValue
        SmValue = Math.min(SmValue,
                  Math.abs(arr[i]));
    }

    // Update MaxAltSum
    MaxAltSum = sum;

    // If cntNeg is
    // an odd number
    if (cntNeg % 2 == 1)
    {

        // Update MaxAltSum
        MaxAltSum -= 2 * SmValue;
    }
    return MaxAltSum;
}

    // Driver Code
    let arr = [ 1, 1, -2, -4, 5 ];
    let N = arr.length;

    document.write(findMaxSumByAlternatingSign(
    arr, N));

// This code is contributed by souravghosh0416.
</script>

Output:

13

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