Modify array to maximize sum of adjacent differences

Given an array, we need to modify values of this array in such a way that sum of absolute differences between two consecutive elements is maximized. If the value of an array element is X, then we can change it to either 1 or X.
Examples :

Input  : arr[] = [3, 2, 1, 4, 5]
Output : 8
We can modify above array as,
Modified arr[] = [3, 1, 1, 4, 1]
Sum of differences = 
|1-3| + |1-1| + |4-1| + |1-4| = 8
Which is the maximum obtainable value 
among all choices of modification.

Input  : arr[] = [1, 8, 9]
Output : 14

This problem is a variation of Assembly Line Scheduling and can be solved using dynamic programming. We need to maximize sum of differences each value X should be changed to either 1 or X. To achieve above stated condition we take a dp array of array length size with 2 columns, where dp[i][0] stores the maximum value of sum using first i elements only if ith array value is modified to 1 and dp[i][1] stores the maximum value of sum using first i elements if ith array value is kept as a[i] itself.Main thing to observe is,

C++

//  C++ program to get maximum consecutive element
// difference sum
#include <bits/stdc++.h>
using namespace std;

// Returns maximum-difference-sum with array
// modifications allowed.
int maximumDifferenceSum(int arr[], int N)
{
    // Initialize dp[][] with 0 values.
    int dp[N][2];
    for (int i = 0; i < N; i++)
        dp[i][0] = dp[i][1] = 0;

    for (int i=0; i<(N-1); i++)
    {
        /*  for [i+1][0] (i.e. current modified
            value is 1), choose maximum from
            dp[i][0] + abs(1 - 1) = dp[i][0] and
            dp[i][1] + abs(1 - arr[i])   */
        dp[i + 1][0] = max(dp[i][0],
                          dp[i][1] + abs(1-arr[i]));

        /*  for [i+1][1] (i.e. current modified value
            is arr[i+1]), choose maximum from
            dp[i][0] + abs(arr[i+1] - 1)    and
            dp[i][1] + abs(arr[i+1] - arr[i])*/
        dp[i + 1][1] = max(dp[i][0] + abs(arr[i+1] - 1),
                     dp[i][1] + abs(arr[i+1] - arr[i]));
    }

    return max(dp[N-1][0], dp[N-1][1]);
}

//  Driver code to test above methods
int main()
{
    int arr[] = {3, 2, 1, 4, 5};
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << maximumDifferenceSum(arr, N) << endl;
    return 0;
}

Java

// java program to get maximum consecutive element
// difference sum
import java.io.*;

class GFG 
{
    // Returns maximum-difference-sum with array
    // modifications allowed.
    static int maximumDifferenceSum(int arr[], int N)
    {
        // Initialize dp[][] with 0 values.
        int dp[][] = new int [N][2];

        for (int i = 0; i < N; i++)
            dp[i][0] = dp[i][1] = 0;
    
        for (int i = 0; i< (N - 1); i++)
        {
            /* for [i+1][0] (i.e. current modified
            value is 1), choose maximum from
            dp[i][0] + abs(1 - 1) = dp[i][0] and
            dp[i][1] + abs(1 - arr[i]) */
            dp[i + 1][0] = Math.max(dp[i][0],
                           dp[i][1] + Math.abs(1 - arr[i]));
    
            /* for [i+1][1] (i.e. current modified value
            is arr[i+1]), choose maximum from
            dp[i][0] + abs(arr[i+1] - 1) and
            dp[i][1] + abs(arr[i+1] - arr[i])*/
            dp[i + 1][1] = Math.max(dp[i][0] + 
                           Math.abs(arr[i + 1] - 1),
                           dp[i][1] + Math.abs(arr[i + 1] 
                           - arr[i]));
        }
    
        return Math.max(dp[N - 1][0], dp[N - 1][1]);
    }

    // Driver code 
    public static void main (String[] args) 
    {
        int arr[] = {3, 2, 1, 4, 5};
        int N = arr.length;
        System.out.println( maximumDifferenceSum(arr, N));
                
    }
}

// This code is contributed by vt_m

C#

// C# program to get maximum consecutive element
// difference sum
using System;

class GFG {
    
    // Returns maximum-difference-sum with array
    // modifications allowed.
    static int maximumDifferenceSum(int []arr, int N)
    {
        
        // Initialize dp[][] with 0 values.
        int [,]dp = new int [N,2];

        for (int i = 0; i < N; i++)
            dp[i,0] = dp[i,1] = 0;
    
        for (int i = 0; i < (N - 1); i++)
        {
            /* for [i+1][0] (i.e. current modified
            value is 1), choose maximum from
            dp[i][0] + abs(1 - 1) = dp[i][0] and
            dp[i][1] + abs(1 - arr[i]) */
            dp[i + 1,0] = Math.Max(dp[i,0],
                        dp[i,1] + Math.Abs(1 - arr[i]));
    
            /* for [i+1][1] (i.e. current modified value
            is arr[i+1]), choose maximum from
            dp[i][0] + abs(arr[i+1] - 1) and
            dp[i][1] + abs(arr[i+1] - arr[i])*/
            dp[i + 1,1] = Math.Max(dp[i,0] + 
                        Math.Abs(arr[i + 1] - 1),
                        dp[i,1] + Math.Abs(arr[i + 1] 
                        - arr[i]));
        }
    
        return Math.Max(dp[N - 1,0], dp[N - 1,1]);
    }

    // Driver code 
    public static void Main () 
    {
        int []arr = {3, 2, 1, 4, 5};
        int N = arr.Length;
        
        Console.Write( maximumDifferenceSum(arr, N));
    }
}

// This code is contributed by nitin mittal.

PHP

<?php
// PHP program to get maximum 
// consecutive element 
// difference sum

// Returns maximum-difference-sum 
// with array modifications allowed.
function maximumDifferenceSum($arr, $N)
{
    // Initialize dp[][] 
    // with 0 values.
    $dp = array(array());
    for ($i = 0; $i < $N; $i++)
        $dp[$i][0] = $dp[$i][1] = 0;

    for ($i = 0; $i < ($N - 1); $i++)
    {
        /* for [i+1][0] (i.e. current 
            modified value is 1), choose 
            maximum from dp[$i][0] + 
            abs(1 - 1) = dp[i][0] and 
            dp[$i][1] + abs(1 - arr[i]) */
        $dp[$i + 1][0] = max($dp[$i][0],
                            $dp[$i][1] + 
                            abs(1 - $arr[$i]));

        /* for [i+1][1] (i.e. current 
            modified value is arr[i+1]), 
            choose maximum from dp[i][0] + 
            abs(arr[i+1] - 1) and dp[i][1] + 
            abs(arr[i+1] - arr[i])*/
        $dp[$i + 1][1] = max($dp[$i][0] + 
                             abs($arr[$i + 1] - 1),
                                       $dp[$i][1] + 
                                 abs($arr[$i + 1] - 
                                        $arr[$i]));
    }

    return max($dp[$N - 1][0], $dp[$N - 1][1]);
}

// Driver Code
$arr = array(3, 2, 1, 4, 5);
$N = count($arr);
echo maximumDifferenceSum($arr, $N);

// This code is contributed by anuj_67.
?>



Output :

8

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

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.



Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : nitin mittal, vt_m




Practice Tags :
Article Tags :
Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.

Recommended Posts:



3.4 Average Difficulty : 3.4/5.0
Based on 22 vote(s)






User Actions