在元素上交替放置加减号后最大化子序列和

给定一个大小为N的数组arr[] 。任务是找到通过替代减法可以达到的最大分数 - 加数组中任何子序列的元素。

例子：

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

方法：我们可以使用动态规划方法来解决这个问题。

• 假设dp1 _i ⇢ 是前i个元素的前缀上的子序列的最大可能总和，前提是子序列的长度是奇数。同样输入dp2 _i ⇢ 仅用于偶数长度的子序列。
• 然后dp1 _i和dp2 _i很容易重新计算为：
  • dp1 _i+1 = max(dp1 _i , dp2 _i +arr _i )
  • dp2 _i+1 = max(dp2 _i , dp1 _i −arr _i )
• 初始值为dp1 ₀ = -∞ , dp2 ₀ = 0并且答案将存储在max ( dp1 _n , dp2 _n ) 中。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
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 array
int maxScore(int arr[], int N)
{
    vector dp1(N + 1);
    vector 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 Code
int 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 approach
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(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 approach
INF = 1e9
 
# Function to find the maximum score
# achieved by alternative minus-plus
# of elements of a subsequence in
# the given array
def 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 Code
arr = [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 approach
using 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

 
        // 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
    

输出

10

时间复杂度： O(N)
辅助空间： O(N)