Given an array arr[] of size N. The task is to find the maximum score that can be achieved by alternative minus – plus of elements of any subsequence in the array.
Example:
Input: arr[] = {2, 3, 6, 5, 4, 7, 8}, N = 7
Output: 10
Explanation: Pick the sequence as {2, 3, 6, 5, 4, 7, 8} = {6, 4, 8}. So, alternative minus – plus = (6 – 4 + 8) = 10.Input: {9, 2, 4, 5, 3}, N =5
Output: 12
Approach: We can use Dynamic Programming method to solve the problem.
- Let dp1i ⇢ be the maximum possible sum of a subsequence on a prefix from the first i elements, provided that the length of the subsequence is odd. Similarly enter dp2i ⇢ only for subsequences of even length.
- Then dp1i and dp2i are easy to recalculate as :
- dp1i+1 = max(dp1i, dp2i+arri)
- dp2i+1 = max(dp2i, dp1i−arri)
- The initial values are dp10 = −∞, dp20 = 0 and the answer will be stored in max(dp1n, dp2n).
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;#define INF 1e9// Function to find the maximum score// achieved by alternative minus-plus// of elements of a subsequence in// the given arrayint maxScore(int arr[], int N){ vector<int> dp1(N + 1); vector<int> dp2(N + 1); dp1[0] = -INF; dp2[0] = 0; for (int i = 0; i < N; ++i) { dp1[i + 1] = max(dp1[i], dp2[i] + arr[i]); dp2[i + 1] = max(dp2[i], dp1[i] - arr[i]); } // Return the maximum return max(dp1.back(), dp2.back());}// Driver Codeint main(){ int arr[] = { 2, 3, 6, 5, 4, 7, 8 }; int N = sizeof(arr) / sizeof(arr[0]); cout << maxScore(arr, N); return 0;} |
Java
// Java program for the above approachclass GFG { static double INF = 1E9; // Function to find the maximum score // achieved by alternative minus-plus // of elements of a subsequence in // the given array public static int maxScore(int arr[], int N) { int[] dp1 = new int[N + 1]; int[] dp2 = new int[N + 1]; dp1[0] = (int) -INF; dp2[0] = 0; for (int i = 0; i < N; ++i) { dp1[i + 1] = Math.max(dp1[i], dp2[i] + arr[i]); dp2[i + 1] = Math.max(dp2[i], dp1[i] - arr[i]); } // Return the maximum return Math.max(dp1[dp1.length - 1], dp2[dp2.length - 1]); } // Driver Code public static void main(String args[]) { int arr[] = { 2, 3, 6, 5, 4, 7, 8 }; int N = arr.length; System.out.println(maxScore(arr, N)); }}// This code is contributed by saurabh_jaiswal. |
Python3
# Python Program to implement# the above approachINF = 1e9# Function to find the maximum score# achieved by alternative minus-plus# of elements of a subsequence in# the given arraydef maxScore(arr, N): dp1 = [0] * (N + 1) dp2 = [0] * (N + 1) dp1[0] = -INF dp2[0] = 0 for i in range(N): dp1[i + 1] = max(dp1[i], dp2[i] + arr[i]) dp2[i + 1] = max(dp2[i], dp1[i] - arr[i]) # Return the maximum return max(dp1[len(dp1) - 1], dp2[len(dp2) - 1])# Driver Codearr = [2, 3, 6, 5, 4, 7, 8]N = len(arr)print(maxScore(arr, N))# This code is contributed by Saurabh Jaiswal |
C#
// C# code to implement the above approachusing System;class GFG { static double INF = 1E9; // Function to find the maximum score // achieved by alternative minus-plus // of elements of a subsequence in // the given array public static int maxScore(int []arr, int N) { int []dp1 = new int[N + 1]; int []dp2 = new int[N + 1]; dp1[0] = (int) -INF; dp2[0] = 0; for (int i = 0; i < N; ++i) { dp1[i + 1] = Math.Max(dp1[i], dp2[i] + arr[i]); dp2[i + 1] = Math.Max(dp2[i], dp1[i] - arr[i]); } // Return the maximum return Math.Max(dp1[dp1.Length - 1], dp2[dp2.Length - 1]); } // Driver Code public static void Main() { int []arr = { 2, 3, 6, 5, 4, 7, 8 }; int N = arr.Length; Console.Write(maxScore(arr, N)); }}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript Program to implement // the above approach let INF = 1e9 // Function to find the maximum score // achieved by alternative minus-plus // of elements of a subsequence in // the given array function maxScore(arr, N) { let dp1 = new Array(N + 1); let dp2 = new Array(N + 1); dp1[0] = -INF; dp2[0] = 0; for (let i = 0; i < N; ++i) { dp1[i + 1] = Math.max(dp1[i], dp2[i] + arr[i]); dp2[i + 1] = Math.max(dp2[i], dp1[i] - arr[i]); } // Return the maximum return Math.max(dp1[dp1.length - 1], dp2[dp2.length - 1]); } // Driver Code let arr = [2, 3, 6, 5, 4, 7, 8] let N = arr.length; document.write(maxScore(arr, N)); // This code is contributed by Potta Lokesh </script> |
Output
10
Time Complexity: O(N)
Auxiliary Space: O(N)
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