Given an array, we need to modify the values of this array in such a way that the 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
Method 1: 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 methodsint 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 sumimport 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 |
Python3
# Python3 program to get maximum # consecutive element difference sum # Returns maximum-difference-sum # with array modifications allowed. def maximumDifferenceSum(arr, N): # Initialize dp[][] with 0 values. dp = [[0, 0] for i in range(N)] for i in range(N): dp[i][0] = dp[i][1] = 0 for i in range(N - 1): # 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 Codeif __name__ == '__main__': arr = [3, 2, 1, 4, 5] N = len(arr) print(maximumDifferenceSum(arr, N))# This code is contributed by PranchalK |
C#
// C# program to get maximum consecutive element// difference sumusing 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.?> |
Javascript
<script> // Javascript 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. let dp = new Array(N); for (let i = 0; i < N; i++) { dp[i] = new Array(2); for (let j = 0; j < 2; j++) { dp[i][j] = 0; } } for (let i = 0; i < N; i++) dp[i][0] = dp[i][1] = 0; for (let 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]); } let arr = [3, 2, 1, 4, 5]; let N = arr.length; document.write( maximumDifferenceSum(arr, N)); // This code is contributed by rameshtravel07.</script> |
Output
8
Time Complexity : O(N)
Auxiliary Space : O(N)
This article is contributed by Utkarsh Trivedi.
Method 2:
For calculating answers at any moment, only previous state values are needed. So instead of using the entire DP array, we can reduce space complexity by using just two variables prev_change and prev_nochange.
C++
// C++ program for the above approach#include <iostream>using namespace std;int maximumDifferenceSum(int arr[], int n) { int prev_change = 0, prev_nochange = 0; for (int i = 1; i < n; i++) { int change = max(prev_change, arr[i - 1] - 1 + prev_nochange); int nochange = max(prev_nochange + abs(arr[i] - arr[i - 1]), arr[i] - 1 + prev_change); prev_change = change; prev_nochange = nochange; } return max(prev_change, prev_nochange);}int main() { int arr[] = {3, 2, 1, 4, 5}; int N = sizeof(arr) / sizeof(arr[0]); cout << maximumDifferenceSum(arr, N) << endl; return 0;}// This code is contributed by lokeshpotta20. |
Java
import java.io.*;import java.util.*; class GFG { static int maximumDifferenceSum(int arr[], int n) { int prev_change = 0, prev_nochange = 0; for (int i = 1; i < n; i++) { int change = Math.max(prev_change, arr[i - 1] - 1 + prev_nochange); int nochange = Math.max(prev_nochange + Math.abs(arr[i] - arr[i - 1]), arr[i] - 1 + prev_change); prev_change = change; prev_nochange = nochange; } return Math.max(prev_change, prev_nochange); } public static void main(String args[]) { int arr[] = {3, 2, 1, 4, 5}; int N = arr.length; System.out.println(maximumDifferenceSum(arr, N)); }} |
Python3
def maximumDifferenceSum(arr, n): prev_change, prev_nochange = 0,0 for i in range(1,n): change = max(prev_change , arr[i-1] -1 + prev_nochange) nochange = max(prev_nochange +abs(arr[i] - arr[i-1]) , arr[i]-1 + prev_change) prev_change = change prev_nochange = nochange return max(prev_change, prev_nochange) if __name__ == '__main__': arr = [3, 2, 1, 4, 5] N = len(arr) print(maximumDifferenceSum(arr, N)) |
C#
// C# program for the above approachusing System;using System.Linq;using System.Collections.Generic;class GFG { static int maximumDifferenceSum(int[] arr, int n) { int prev_change = 0, prev_nochange = 0; for (int i = 1; i < n; i++) { int change = Math.Max(prev_change, arr[i - 1] - 1 + prev_nochange); int nochange = Math.Max(prev_nochange + Math.Abs(arr[i] - arr[i - 1]), arr[i] - 1 + prev_change); prev_change = change; prev_nochange = nochange; } return Math.Max(prev_change, prev_nochange); } static public void Main() { int[] arr = {3, 2, 1, 4, 5}; int N = arr.Length; Console.Write(maximumDifferenceSum(arr, N)); }}// This code is contributed by ratiagrawal. |
Javascript
// Javascript program for the above approachfunction maximumDifferenceSum( arr, n) { let prev_change = 0, prev_nochange = 0; for (let i = 1; i < n; i++) { let change = Math.max(prev_change, arr[i - 1] - 1 + prev_nochange); let nochange = Math.max(prev_nochange + Math.abs(arr[i] - arr[i - 1]), arr[i] - 1 + prev_change); prev_change = change; prev_nochange = nochange; } return Math.max(prev_change, prev_nochange);}let arr = [3, 2, 1, 4, 5];let N = arr.length;console.log(maximumDifferenceSum(arr, N)); |
Output
8
Time Complexity: O(N)
Auxiliary Space: O(1)
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