Minimum cost to convert given string to consist of only vowels

Given string str of lower case alphabets, the task is to find the minimum cost to change the input string in a string that contains only vowels. Each consonant is changed to the nearest vowels. The cost is defined as the absolute difference between the ASCII value of consonant and vowel.

Examples:

Input: str = “abcde”
Output: 4
Explanation:
Here, a and e are already the vowels but b, c, and d are consonants. So b, c, and d are changed to nearest vowels as:
1. b –> a  = |98 – 97| = 1, 
2. c –> a  = |99 – 97| = 2  or c –> e  = |99 – 101| = 2 ( to minimum the cost), 
3. d –> e  = |100 – 101| = 1.
Therefore, the minimum cost is 1 + 2 + 1 = 4.

Input: str = “aaa” 
Output: 0
Explanation: 
There is no consonant in the string.

 

Approach: The idea is to traverse over the string and replace each consonant with their nearest vowel and add the cost as the difference between the consonants and changed vowel. Print the cost after all operations.



Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for the above approach 
#include<bits/stdc++.h>
using namespace std;
  
// Function to find the minimum cost 
int min_cost(string st)
{
      
    // Store vowels 
    string vow = "aeiou";
    int cost = 0;
  
    // Loop for iteration of string 
    for(int i = 0; i < st.size(); i++)
    {
        vector<int> costs; 
  
        // Loop to calculate the cost 
        for(int j = 0; j < 5; j++) 
            costs.push_back(abs(st[i] - vow[j])); 
              
        // Add minimum cost 
        cost += *min_element(costs.begin(), 
                             costs.end()); 
    }
    return cost;
}
  
// Driver Code 
int main()
{
      
    // Given String 
    string str = "abcde";
      
    // Function Call 
    cout << (min_cost(str));
}
  
// This code is contributed by grand_master

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program for above approach
  
# Function to find the minimum cost
def min_cost(st):
  
    # Store vowels
    vow = "aeiou"
    cost = 0
  
    # Loop for iteration of string
    for i in range(len(st)):
        costs = []
  
        # Loop to calculate the cost
        for j in range(5):
            costs.append(abs(ord(st[i])-ord(vow[j])))
              
        # Add minimum cost
        cost += min(costs)
    return cost
  
# Driver Code
  
# Given String
str = "abcde"
  
# Function Call
print(min_cost(str))

chevron_right


Output:

4

Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : grand_master