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++ 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> |
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.