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 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 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
<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>
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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