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> |
Output:
8
Time Complexity: O(N)
Auxiliary Space: O(26)
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