使字符串不含子序列的最小成本

给定由小写英文字母组成的字符串str和一个长度相同的正整数arr[]数组。任务是从给定的字符串，该字符串中没有子序列形成的字符串“代码”除去一些字符。删除字符str[i] 的成本是arr[i] 。找出实现目标的最低成本。

例子：

Input: str = “code”, arr[] = {3, 2, 1, 3} 
Output: 1 
Remove ‘d’ which costs the minimum.
Input: str = “ccooddde”, arr[] = {3, 2, 1, 3, 3, 5, 1, 6} 
Output: 4 
Remove both the ‘o’ which cost 1 + 3 = 4 

方法：如果任何带有“代码”的子序列是可能的，则需要删除单个字符。删除每个字符的成本在arr[] 中给出。因此，遍历字符串并为c 、 o 、 d或e 的每个字符计算删除它们的成本。最后，移除所有字符的成本中的最小值是所需的最小成本。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the minimum cost
int findCost(string str, int arr[], int n)
{
    long long costofC = 0, costofO = 0,
              costofD = 0, costofE = 0;
 
    // Traverse the string
    for (int i = 0; i < n; i++) {
 
        // Min Cost to remove 'c'
        if (str[i] == 'c')
            costofC += arr[i];
 
        // Min Cost to remove subsequence "co"
        else if (str[i] == 'o')
            costofO = min(costofC, costofO + arr[i]);
 
        // Min Cost to remove subsequence "cod"
        else if (str[i] == 'd')
            costofD = min(costofO, costofD + arr[i]);
 
        // Min Cost to remove subsequence "code"
        else if (str[i] == 'e')
            costofE = min(costofD, costofE + arr[i]);
    }
 
    // Return the minimum cost
    return costofE;
}
 
// Driver program
int main()
{
    string str = "geekcodergeeks";
    int arr[] = { 1, 2, 1, 3, 4, 2, 6, 4, 6, 2, 3, 3, 3, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findCost(str, arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
 
class GFG {
 
    // Function to return the minimum cost
    static int findCost(String str, int arr[], int n)
    {
        long costofC = 0, costofO = 0,
             costofD = 0, costofE = 0;
 
        // Traverse the string
        for (int i = 0; i < n; i++) {
 
            // Min Cost to remove 'c'
            if (str.charAt(i) == 'c')
                costofC += arr[i];
 
            // Min Cost to remove subsequence "co"
            else if (str.charAt(i) == 'o')
                costofO = Math.min(costofC, costofO + arr[i]);
 
            // Min Cost to remove subsequence "cod"
            else if (str.charAt(i) == 'd')
                costofD = Math.min(costofO, costofD + arr[i]);
 
            // Min Cost to remove subsequence "code"
            else if (str.charAt(i) == 'e')
                costofE = Math.min(costofD, costofE + arr[i]);
        }
 
        // Return the minimum cost
        return (int)costofE;
    }
 
    // Driver program
    public static void main(String[] args)
    {
        String str = "geekcodergeeks";
        int arr[] = { 1, 2, 1, 3, 4, 2, 6, 4, 6, 2, 3, 3, 3, 2 };
        int n = arr.length;
        System.out.print(findCost(str, arr, n));
    }
}
// This code has been contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
# Function to return the minimum cost
def findCost(str, arr, n):
    costofC, costofO = 0, 0
    costofD, costofE = 0, 0
 
    # Traverse the string
    for i in range(n):
 
        # Min Cost to remove 'c'
        if (str[i] == 'c'):
            costofC += arr[i]
 
        # Min Cost to remove subsequence "co"
        elif (str[i] == 'o'):
            costofO = min(costofC, costofO + arr[i])
 
        # Min Cost to remove subsequence "cod"
        elif (str[i] == 'd'):
            costofD = min(costofO, costofD + arr[i])
 
        # Min Cost to remove subsequence "code"
        elif (str[i] == 'e'):
            costofE = min(costofD, costofE + arr[i])
 
    # Return the minimum cost
    return costofE
 
# Driver Code
if __name__ == '__main__':
    str = "geekcodergeeks"
    arr = [1, 2, 1, 3, 4, 2, 6, 4, 6, 2, 3, 3, 3, 2]
    n = len(arr)
    print(findCost(str, arr, n))
 
# This code contributed by PrinciRaj1992

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the minimum cost
    public static int findCost(string str,
                            int[] arr, int n)
    {
        long costofC = 0, costofO = 0,
              costofD = 0, costofE = 0;
 
        // Traverse the string
        for (int i = 0; i < n; i++)
        {
 
            // Min Cost to remove 'c'
            if (str[i] == 'c')
            {
                costofC += arr[i];
            }
 
            // Min Cost to remove subsequence "co"
            else if (str[i] == 'o')
            {
                costofO = Math.Min(costofC, costofO + arr[i]);
            }
 
            // Min Cost to remove subsequence "cod"
            else if (str[i] == 'd')
            {
                costofD = Math.Min(costofO, costofD + arr[i]);
            }
 
            // Min Cost to remove subsequence "code"
            else if (str[i] == 'e')
            {
                costofE = Math.Min(costofD, costofE + arr[i]);
            }
        }
 
        // Return the minimum cost
        return (int)costofE;
    }
 
    // Driver program
    public static void Main(string[] args)
    {
        string str = "geekcodergeeks";
        int[] arr = new int[] {1, 2, 1, 3, 4, 2, 6,
                                4, 6, 2, 3, 3, 3, 2};
        int n = arr.Length;
        Console.Write(findCost(str, arr, n));
    }
}
 
// This code is contributed by shrikanth13

Javascript

 
 
// Javascript implementation of the approach
 
// Function to return the Math.minimum cost
function findCost(str, arr, n)
{
    var costofC = 0, costofO = 0,
              costofD = 0, costofE = 0;
 
    // Traverse the string
    for (var i = 0; i < n; i++) {
 
        // Math.min Cost to remove 'c'
        if (str[i] == 'c')
            costofC += arr[i];
 
        // Math.min Cost to remove subsequence "co"
        else if (str[i] == 'o')
            costofO = Math.min(costofC, costofO + arr[i]);
 
        // Math.min Cost to remove subsequence "cod"
        else if (str[i] == 'd')
            costofD = Math.min(costofO, costofD + arr[i]);
 
        // Math.min Cost to remove subsequence "code"
        else if (str[i] == 'e')
            costofE = Math.min(costofD, costofE + arr[i]);
    }
 
    // Return the Math.minimum cost
    return costofE;
}
 
// Driver program
var str = "geekcodergeeks";
var arr = [ 1, 2, 1, 3, 4, 2, 6, 4, 6, 2, 3, 3, 3, 2 ];
var n = arr.length;
document.write( findCost(str, arr, n));
 

输出：

2

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