Maximum possible array sum after performing the given operation

Given an array arr[] of size N, the task is to find the maximum sum of the elements of the array after applying the given operation any number of times. In a single operation, choose an index 1 ? i < N and multiply both arr[i] and arr[i - 1] by -1.
Examples: 

Input: arr[] = {-10, 5, -4} 
Output: 19 
Perform the operation for i = 1 and 
the array becomes {10, -5, -4} 
Perform the operation for i = 2 and 
the array becomes {10, 5, 4} 
10 + 5 + 4 = 19
Input: arr[] = {10, -4, -8, -11, 3} 
Output: 30 

Approach: This problem can be solved using dynamic programming. Since it is useless to choose and flip the same arr[i], we will consider flipping at most once for each element of the array in order from the left. Let dp[i][0] represents the maximum possible sum up to the i^th index without flipping the i^th index. dp[i][1] represents the maximum possible sum up to the i^th index with flipping the i^th index. So, dp(n, 0) is our required answer. 
Base conditions: 

1. dp[0][0] = 0
2. dp[0][1] = INT_MIN

Recurrence relation:  

1. dp[i + 1][0] = max(dp[i][0] + arr[i], dp[i][1] - arr[i])
2. dp[i + 1][1] = max(dp[i][0] - arr[i], dp[i][1] + arr[i])

Below is the implementation of the above approach:  

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the maximum possible
// sum after performing the given operation
int max_sum(int a[], int n)
{
    // Dp vector to store the answer
    vector<vector<int> > dp(n + 1,
                            vector<int>(2, 0));

    // Base value
    dp[0][0] = 0, dp[0][1] = -999999;

    for (int i = 0; i <= n - 1; i++) {
        dp[i + 1][0] = max(dp[i][0] + a[i],
                           dp[i][1] - a[i]);
        dp[i + 1][1] = max(dp[i][0] - a[i],
                           dp[i][1] + a[i]);
    }

    // Return the maximum sum
    return dp[n][0];
}

// Driver code
int main()
{
    int a[] = { -10, 5, -4 };
    int n = sizeof(a) / sizeof(a[0]);

    cout << max_sum(a, n);

    return 0;
}

// Java implementation of the approach
class GFG 
{

// Function to return the maximum possible
// sum after performing the given operation
static int max_sum(int a[], int n)
{
    // Dp vector to store the answer
    int [][]dp = new int[n + 1][2];
                        
    // Base value
    dp[0][0] = 0; dp[0][1] = -999999;

    for (int i = 0; i <= n - 1; i++) 
    {
        dp[i + 1][0] = Math.max(dp[i][0] + a[i],
                                dp[i][1] - a[i]);
        dp[i + 1][1] = Math.max(dp[i][0] - a[i],
                                dp[i][1] + a[i]);
    }

    // Return the maximum sum
    return dp[n][0];
}

// Driver code
public static void main(String[] args) 
{
    int a[] = { -10, 5, -4 };
    int n = a.length;

    System.out.println(max_sum(a, n));
}
}

// This code is contributed by 29AjayKumar

# Python implementation of the approach

# Function to return the maximum possible
# sum after performing the given operation
def max_sum(a, n):
    # Dp vector to store the answer
    dp = [[0 for i in range(2)] for j in range(n+1)]
                        
    # Base value
    dp[0][0] = 0; dp[0][1] = -999999;

    for i in range(0, n): 
        dp[i + 1][0] = max(dp[i][0] + a[i],
                                dp[i][1] - a[i]);
        dp[i + 1][1] = max(dp[i][0] - a[i],
                                dp[i][1] + a[i]);

    # Return the maximum sum
    return dp[n][0];

# Driver code
if __name__ == '__main__':
    a = [-10, 5, -4 ];
    n = len(a);

    print(max_sum(a, n));
    
# This code is contributed by 29AjayKumar

<script>
// javascript implementation of the approach    // Function to return the maximum possible
    // sum after performing the given operation
    function max_sum(a , n) {
        // Dp vector to store the answer
        var dp = Array(n + 1).fill().map(()=>Array(2).fill(0));

        // Base value
        dp[0][0] = 0;
        dp[0][1] = -999999;

        for (i = 0; i <= n - 1; i++) {
            dp[i + 1][0] = Math.max(dp[i][0] + a[i], dp[i][1] - a[i]);
            dp[i + 1][1] = Math.max(dp[i][0] - a[i], dp[i][1] + a[i]);
        }

        // Return the maximum sum
        return dp[n][0];
    }

    // Driver code
    
        var a = [ -10, 5, -4 ];
        var n = a.length;

        document.write(max_sum(a, n));

// This code is contributed by todaysgaurav
</script>

// C# implementation of the approach
using System;
    
class GFG 
{

// Function to return the maximum possible
// sum after performing the given operation
static int max_sum(int []a, int n)
{
    // Dp vector to store the answer
    int [,]dp = new int[n + 1, 2];
                        
    // Base value
    dp[0, 0] = 0; dp[0, 1] = -999999;

    for (int i = 0; i <= n - 1; i++) 
    {
        dp[i + 1, 0] = Math.Max(dp[i, 0] + a[i],
                                dp[i, 1] - a[i]);
        dp[i + 1, 1] = Math.Max(dp[i, 0] - a[i],
                                dp[i, 1] + a[i]);
    }

    // Return the maximum sum
    return dp[n, 0];
}

// Driver code
public static void Main(String[] args) 
{
    int []a = { -10, 5, -4 };
    int n = a.Length;

    Console.WriteLine(max_sum(a, n));
}
}

// This code is contributed by PrinciRaj1992

19

Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time.
Auxiliary Space: O(N), as we are using  extra space for DP.

Efficient approach : Space Optimization O(1)

To optimize the space complexity in previous code we are only using the previous row of the dp table to compute the values of the current row. Therefore, we can maintain just two variables to keep track of the values of the previous row instead of using a 2D vector. This will reduce the space complexity from O(N) to O(1).

Implementation Steps:

• Initialize two variables prev0 and prev1 to 0 and -999999 respectively.
• Loop through the input array from index 0 to n-1.
• Compute two new variables curr0 and curr1 as follows.
  • curr0 is the maximum of (prev0 + a[i]) and (prev1 - a[i])      
  • curr1 is the maximum of (prev0 - a[i]) and (prev1 + a[i])
• Update prev0 to curr0 and prev1 to curr1 for further iteration.
• At last return final answer stored in prev0.

Implementation:

// C++ code for above approach

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

// Function to return the maximum possible
// sum after performing the given operation
int max_sum(int a[], int n)
{
    // Variables to store the values of the
    // previous row of the DP table
    int prev0 = 0, prev1 = -999999;

    for (int i = 0; i <= n - 1; i++) {
        int curr0 = max(prev0 + a[i], prev1 - a[i]);
        int curr1 = max(prev0 - a[i], prev1 + a[i]);
      
          // assigning values for further iterations
        prev0 = curr0;
        prev1 = curr1;
    }

    // Return the maximum sum
    return prev0;
}

// Driver code
int main()
{
    int a[] = { -10, 5, -4 };
    int n = sizeof(a) / sizeof(a[0]);

    cout << max_sum(a, n);

    return 0;
}

// Java code for above approach

import java.util.*;

class Main {
  // Function to return the maximum possible
  // sum after performing the given operation
  static int maxSum(int[] a, int n)
  {
      // Variables to store the values of the
      // previous row of the DP table
      int prev0 = 0, prev1 = -999999;

      for (int i = 0; i <= n - 1; i++) {
          int curr0 = Math.max(prev0 + a[i], prev1 - a[i]);
          int curr1 = Math.max(prev0 - a[i], prev1 + a[i]);

          // assigning values for further iterations
          prev0 = curr0;
          prev1 = curr1;
      }

      // Return the maximum sum
      return prev0;
  }

  // Driver code
  public static void main(String[] args)
  {
      int[] a = { -10, 5, -4 };
      int n = a.length;

      System.out.println(maxSum(a, n));
  }
}

# Function to return the maximum possible sum after performing the given operation
def max_sum(a, n):
    # Variables to store the values of the previous row of the DP table
    prev0 = 0
    prev1 = -999999

    for i in range(n):
        curr0 = max(prev0 + a[i], prev1 - a[i])
        curr1 = max(prev0 - a[i], prev1 + a[i])
      
        # assigning values for further iterations
        prev0 = curr0
        prev1 = curr1

    # Return the maximum sum
    return prev0

# Driver code
a = [-10, 5, -4]
n = len(a)

print(max_sum(a, n))

using System;

class MainClass
{
    // Function to return the maximum possible
    // sum after performing the given operation
    static int maxSum(int[] a, int n)
    {
        // Variables to store the values of the
        // previous row of the DP table
        int prev0 = 0, prev1 = -999999;

        for (int i = 0; i <= n - 1; i++)
        {
            int curr0 = Math.Max(prev0 + a[i], prev1 - a[i]);
            int curr1 = Math.Max(prev0 - a[i], prev1 + a[i]);

            // assigning values for further iterations
            prev0 = curr0;
            prev1 = curr1;
        }

        // Return the maximum sum
        return prev0;
    }

    // Driver code
    public static void Main()
    {
        int[] a = { -10, 5, -4 };
        int n = a.Length;

        Console.WriteLine(maxSum(a, n));
    }
}

// Function to return the maximum possible
// sum after performing the given operation
function max_sum(a, n) {
    // Variables to store the values of the
    // previous row of the DP table
    let prev0 = 0,
        prev1 = -999999;

    for (let i = 0; i <= n - 1; i++) {
        let curr0 = Math.max(prev0 + a[i], prev1 - a[i]);
        let curr1 = Math.max(prev0 - a[i], prev1 + a[i]);

        // assigning values for further iterations
        prev0 = curr0;
        prev1 = curr1;
    }

    // Return the maximum sum
    return prev0;
}

// Driver code
let a = [-10, 5, -4];
let n = a.length;

console.log(max_sum(a, n));

19

Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time.
Auxiliary Space: O(1)

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Article Tags :

• Dynamic Programming
• DSA
• Arrays

Practice Tags :

• Arrays
• Dynamic Programming