Minimum cost to make a string free of a subsequence

Last Updated : 15 Dec, 2022

Given string str consisting of lowercase English alphabets and an array of positive integer arr[] both of the same length. The task is to remove some characters from the given string such that no sub-sequence in the string forms the string "code". Cost of removing a character str[i] is arr[i]. Find the minimum cost to achieve the target.

Examples:

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

Approach: If any sub-sequence with "code" is possible, then the removal of a single character is required. Cost for the removal of each character is given in arr[]. So, traverse the string and for each character which is either c, o, d or e calculate the cost of their removal. And finally, the minimum among the cost of removal of all characters is required minimum cost.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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# 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

<script>


// 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));

</script>

Output:

Time Complexity: O(n), where n is the size of the given array and string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Analysis of Algorithms

Shivam.Pradhan

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