Open In App

Minimize cost to empty a given string by removing characters alphabetically

Improve
Improve
Like Article
Like
Save
Share
Report

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:
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:

  1. Initialize the total cost to 0.
  2. 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.
  3. 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.
  4. Else subtract the count from its current index and then add it to the total cost.
  5. 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)



Last Updated : 19 Oct, 2021
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads