通过每次移动两次向前跳跃或一次向后跳跃来最小化到达数组末端的成本

给定一个由N 个正整数组成的数组arr[] ，任务是通过仅移动到索引(i + 2)和(i – 1)来找到穿过数组或到达数组末尾所需的最小成本第i^个索引。

例子：

Input: arr[] = {5, 1, 2, 10, 100}
Output: 18
Explanation:
Optimal cost path (0 based indexing): 0 → 2 → 1 → 3 → 5
Therefore, the minimum cost = 5 + 2 + 1 + 10 = 18.

Input: arr[] = {9, 4, 6, 8, 5}
Output: 20
Explanation:
Optimal cost path (0 based indexing): 0 → 2 → 4
Therefore, the minimum cost = 9 + 6 + 5 = 20

朴素的方法：可以根据以下观察解决给定的问题：

• 由于所有成本都是正的，因此向后移动多于一步永远不是最佳选择，因此要到达数组的特定索引i ，要么直接从第(i – 2)^个索引跳转，要么从(i – 1) ^th到(i + 1) ^th index，即(向前跳2次)，后跟向后跳1^次，即从(i + 1) ^th index到^第i index。
• 现在，从数组的末尾递归遍历索引(i – 2)和(i – 1)处的元素，计算两者的最小成本。因此，可以使用以下递推关系计算交叉阵列的最小成本：

minCost(index) = minimum(minCost(index – 2) + arr[i], minCost(index – 1) + arr[i] + arr[i + 1])

时间复杂度： O(2 ^N )
辅助空间： O(1)

高效的方法：上面讨论的方法有最优子结构和重叠子问题。因此，它可以通过使用 Memoization 或 Tabulation 进行优化。请按照以下步骤解决问题：

• 初始化一个数组dp[] ，其中dp[i]存储到达第i^个索引的最小成本。
• 将dp[0] = arr[0]初始化为到达第0^个索引的成本，它等于第0^个索引本身的值。更新DP [1] = ARR [0] + ARR [1] + ARR [2]，作为达到第^1个索引，跳从^第0索引^第二索引indexn给^第一索引。
• 使用变量i迭代范围[2, N – 2]并将dp[i]更新为(dp[i – 2] + arr[i])和(dp[i – 1] + arr[i] ] + arr[i + 1]) 。
• 对于最后一个索引(N – 1) ，将dp[N – 1]更新为(dp[N – 3] + arr[N – 1])和(dp[N – 2]) 的最小值。
• 完成以上步骤后，打印dp[N-1]的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the minimum cost
// to reach the end of an array
void minCost(int arr[], int n)
{
    // Base Case: When N < 3
    if (n < 3) {
        cout << arr[0];
        return;
    }
 
    // Store the results in table
    int* dp = new int[n];
 
    // Initialize base cases
    dp[0] = arr[0];
    dp[1] = dp[0] + arr[1] + arr[2];
 
    // Iterate over the range[2, N - 2]
    // to construct the dp array
    for (int i = 2; i < n - 1; i++)
        dp[i] = min(dp[i - 2] + arr[i],
                    dp[i - 1] + arr[i]
                        + arr[i + 1]);
 
    // Handle case for the last index, i.e. N - 1
    dp[n - 1] = min(dp[n - 2],
                    dp[n - 3] + arr[n - 1]);
 
    // Print the answer
    cout << dp[n - 1];
}
 
// Driver Code
int main()
{
    int arr[] = { 9, 4, 6, 8, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    minCost(arr, N);
 
    return 0;
}

Java

// Java Program to implement
// the above approach
import java.io.*;
import java.util.*;
 
class GFG
{
 
    // Function to find the minimum cost
    // to reach the end of an array
    static void minCost(int arr[], int n)
    {
       
        // Base Case: When N < 3
        if (n < 3) {
            System.out.println(arr[0]);
            return;
        }
 
        // Store the results in table
        int dp[] = new int[n];
 
        // Initialize base cases
        dp[0] = arr[0];
        dp[1] = dp[0] + arr[1] + arr[2];
 
        // Iterate over the range[2, N - 2]
        // to construct the dp array
        for (int i = 2; i < n - 1; i++)
            dp[i] = Math.min(dp[i - 2] + arr[i],
                           dp[i - 1] + arr[i] + arr[i + 1]);
 
        // Handle case for the last index, i.e. N - 1
        dp[n - 1] = Math.min(dp[n - 2], dp[n - 3] + arr[n - 1]);
 
        // Print the answer
        System.out.println(dp[n - 1]);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 9, 4, 6, 8, 5 };
        int N = arr.length;
        minCost(arr, N);
    }
}
 
// This code is contributed by Kingash.

Python3

# Python 3 program for the above approach
 
# Function to find the minimum cost
# to reach the end of an array
def minCost(arr, n):
 
    # Base Case: When N < 3
    if (n < 3):
        print(arr[0])
        return
 
    # Store the results in table
    dp = [0] * n
 
    # Initialize base cases
    dp[0] = arr[0]
    dp[1] = dp[0] + arr[1] + arr[2]
 
    # Iterate over the range[2, N - 2]
    # to construct the dp array
    for i in range(2, n - 1):
        dp[i] = min(dp[i - 2] + arr[i],
                    dp[i - 1] + arr[i]
                    + arr[i + 1])
 
    # Handle case for the last index, i.e. N - 1
    dp[n - 1] = min(dp[n - 2],
                    dp[n - 3] + arr[n - 1])
 
    # Print the answer
    print(dp[n - 1])
 
# Driver Code
if __name__ == "__main__":
 
    arr = [9, 4, 6, 8, 5]
    N = len(arr)
    minCost(arr, N)
 
    # This code is contributed by ukasp.

C#

// C# Program to implement
// the above approach
using System;
public class GFG
{
 
  // Function to find the minimum cost
  // to reach the end of an array
  static void minCost(int []arr, int n)
  {
 
    // Base Case: When N < 3
    if (n < 3) {
      Console.WriteLine(arr[0]);
      return;
    }
 
    // Store the results in table
    int []dp = new int[n];
 
    // Initialize base cases
    dp[0] = arr[0];
    dp[1] = dp[0] + arr[1] + arr[2];
 
    // Iterate over the range[2, N - 2]
    // to construct the dp array
    for (int i = 2; i < n - 1; i++)
      dp[i] = Math.Min(dp[i - 2] + arr[i],
                       dp[i - 1] + arr[i] + arr[i + 1]);
 
    // Handle case for the last index, i.e. N - 1
    dp[n - 1] = Math.Min(dp[n - 2], dp[n - 3] + arr[n - 1]);
 
    // Print the answer
    Console.WriteLine(dp[n - 1]);
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
    int []arr = { 9, 4, 6, 8, 5 };
    int N = arr.Length;
    minCost(arr, N);
  }
}
 
// This code is contributed by AnkThon

Javascript

 
// Javascript program for the above approach
 
    // Function to find the minimum cost
    // to reach the end of an array
    function minCost(arr, n)
    {
        
        // Base Case: When N < 3
        if (n < 3) {
            document.write(arr[0]);
            return;
        }
  
        // Store the results in table
        let dp = [];
  
        // Initialize base cases
        dp[0] = arr[0];
        dp[1] = dp[0] + arr[1] + arr[2];
  
        // Iterate over the range[2, N - 2]
        // to construct the dp array
        for (let i = 2; i < n - 1; i++)
            dp[i] = Math.min(dp[i - 2] + arr[i],
                           dp[i - 1] + arr[i] + arr[i + 1]);
  
        // Handle case for the last index, i.e. N - 1
        dp[n - 1] = Math.min(dp[n - 2], dp[n - 3] + arr[n - 1]);
  
        // Prlet the answer
        document.write(dp[n - 1]);
    }
 
// Driver Code
 
        let arr = [ 9, 4, 6, 8, 5 ];
        let N = arr.length;
        minCost(arr, N);
 

输出：

20

时间复杂度： O(N)
辅助空间： O(N)

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