Minimize cost to empty a given string by removing characters alphabetically
Given string str, the task is to minimize the total cost to remove all the characters from the string in alphabetical order.
The cost of removing any character at i th index from the string will be i. The indexing is 1-based.
Examples:
Input: str = “abcab”
Output: 8
Explanation:
First char ‘a’ at index 1 is removed, str[] becomes “bcab”,
Then char ‘a’ with index 3 is removed, str[] becomes “bcb”
After that char ‘b’ with index 1 is removed, str[] becomes “cb”,
Then char ‘b’ with index 2 is removed, str[] becomes “c”,
Finally, char ‘c’ is removed.
Total points = 1+3 + 1 + 2 + 1 = 8.Input: str = “def”
Output: 3
Naive Approach: The simplest approach is to remove the smallest character with a smaller index in the string in each step and keep on adding the cost to the total cost. Print the final cost after this operation.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by precomputing for each character, the number of smaller characters preceding it in the given string. Below are the steps:
- Initialize the total cost to 0.
- Transverse the given string and for each character, count the number of characters that are less than the current character and have occurred before it.
- If this count is 0, this means that the current character will be removed at the present index, so add the index of the character to the resultant cost.
- Else subtract the count from its current index and then add it to the total cost.
- Print the total cost after all the above steps.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> #include <vector> using namespace std; // Function to find the minimum cost // required to remove each character // of the string in alphabetical order int minSteps(string str, int N) { int smaller, cost = 0; // Stores the frequency of // characters of the string int f[26] = { 0 }; // Iterate through the string for ( int i = 0; i < N; i++) { int curr_ele = str[i] - 'a' ; smaller = 0; // Count the number of characters // smaller than the present character for ( int j = 0; j <= curr_ele; j++) { if (f[j]) smaller += f[j]; } // If no smaller character // precedes current character if (smaller == 0) cost += (i + 1); else cost += (i - smaller + 1); // Increase the frequency of // the current character f[str[i] - 'a' ]++; } // Return the // total cost return cost; } // Driver Code int main() { // Given string str string str = "abcab" ; int N = str.size(); // Function call cout << minSteps(str, N); return 0; } |
Java
// Java program for // the above approach import java.io.*; class GFG{ // Function to find the minimum cost // required to remove each character // of the string in alphabetical order static int minSteps(String str, int N) { int smaller, cost = 0 ; // Stores the frequency of // characters of the string int f[] = new int [ 26 ]; // Iterate through the string for ( int i = 0 ; i < N; i++) { int curr_ele = str.charAt(i) - 'a' ; smaller = 0 ; // Count the number of characters // smaller than the present character for ( int j = 0 ; j <= curr_ele; j++) { if (f[j] != 0 ) smaller += f[j]; } // If no smaller character // precedes current character if (smaller == 0 ) cost += (i + 1 ); else cost += (i - smaller + 1 ); // Increase the frequency of // the current character f[str.charAt(i) - 'a' ]++; } // Return the // total cost return cost; } // Driver Code public static void main(String[] args) { // Given string str String str = "abcab" ; int N = str.length(); // Function call System.out.println(minSteps(str, N)); } } // This code is contributed by AnkitRai01 |
Python3
# Python3 program for the above approach # Function to find the minimum cost # required to remove each character # of the string in alphabetical order def minSteps( str , N): cost = 0 # Stores the frequency of # characters of the string f = [ 0 ] * 26 # Iterate through the string for i in range (N): curr_ele = ord ( str [i]) - ord ( 'a' ) smaller = 0 # Count the number of characters # smaller than the present character for j in range (curr_ele + 1 ): if (f[j]): smaller + = f[j] # If no smaller character # precedes current character if (smaller = = 0 ): cost + = (i + 1 ) else : cost + = (i - smaller + 1 ) # Increase the frequency of # the current character f[ ord ( str [i]) - ord ( 'a' )] + = 1 # Return the total cost return cost # Driver Code # Given string str str = "abcab" N = len ( str ) # Function call print (minSteps( str , N)) # This code is contributed by Shivam Singh |
C#
// C# program for // the above approach using System; class GFG{ // Function to find the minimum cost // required to remove each character // of the string in alphabetical order static int minSteps( string str, int N) { int smaller, cost = 0; // Stores the frequency of // characters of the string int [] f = new int [26]; // Iterate through the string for ( int i = 0; i < N; i++) { int curr_ele = str[i] - 'a' ; smaller = 0; // Count the number of characters // smaller than the present character for ( int j = 0; j <= curr_ele; j++) { if (f[j] != 0) smaller += f[j]; } // If no smaller character // precedes current character if (smaller == 0) cost += (i + 1); else cost += (i - smaller + 1); // Increase the frequency of // the current character f[str[i] - 'a' ]++; } // Return the // total cost return cost; } // Driver Code public static void Main() { // Given string str string str = "abcab" ; int N = str.Length; // Function call Console.Write(minSteps(str, N)); } } // This code is contributed by Chitranayal |
Javascript
<script> // JavaScript program for // the above approach // Function to find the minimum cost // required to remove each character // of the string in alphabetical order function minSteps(str, N) { var smaller, cost = 0; // Stores the frequency of // characters of the string var f = new Array(26).fill(0); // Iterate through the string for ( var i = 0; i < N; i++) { var curr_ele = str[i].charCodeAt(0) - "a" .charCodeAt(0); smaller = 0; // Count the number of characters // smaller than the present character for ( var j = 0; j <= curr_ele; j++) { if (f[j] !== 0) smaller += f[j]; } // If no smaller character // precedes current character if (smaller === 0) cost += i + 1; else cost += i - smaller + 1; // Increase the frequency of // the current character f[str[i].charCodeAt(0) - "a" .charCodeAt(0)]++; } // Return the // total cost return cost; } // Driver Code // Given string str var str = "abcab" ; var N = str.length; // Function call document.write(minSteps(str, N)); // This code is contributed by rdtank </script> |
8
Time Complexity: O(N)
Auxiliary Space: O(26)