Modify array to maximize sum of adjacent differences

3.4

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++ 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;
}

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.

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