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 costint 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 programint main(){ string str = "geekcoderneveropen"; 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 approachclass 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 = "geekcoderneveropen"; 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 costdef 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 Codeif __name__ == '__main__': str = "geekcoderneveropen" 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 approachusing 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 = "geekcoderneveropen"; 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 costfunction 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 programvar str = "geekcoderneveropen";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:
2
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.
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