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Minimize cost to replace all the vowels of a given String by a single vowel
  • Last Updated : 31 Jul, 2020

Given a string S, the task is to find the minimum cost to convert all the vowels {a, e, i, o, u} in the given string to any one of them. The cost of converting a vowel is given by the change in ASCII value.
Examples: 
 

Input: S = “geeksforgeeks” 
Output: 10 
Explanation: 
Count of e’s = 4 
Count of o’s = 1 
Conversion from ‘o’ to ‘e’ costs abs(‘o’ – ‘e’) = 10. 
Hence, the output is 10.
Input: S = “coding” 
Output:
Explanation: 
Conversion from ‘o’ to ‘i’ costs abs(‘o’ – ‘i’) = 6. 
Hence, the output is 10. 
 

 

Approach: 
The idea is to calculate the separate cost for converting every vowel in the string to one of the vowels {a, e, i, o, u} and choose the vowel which gives the minimum cost. Follow the steps below: 
 

  • Initialize 5 variables cA, cE, cI, cO, and cU with a cost of 0 which stores the cost of vowels {a, e, i, o, u} respectively.
  • Iterate through the string, and for each vowel find the cost to convert it into all the other vowels and store it in the variables cA, cE, cI, cO, and cU respectively.
  • The minimum cost to convert all the vowels into a single one is given by min(cA, cE, cI, cO, cU).

Below is the implementation of the above approach:
 



C++




// C++ program for the above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function that return true if the 
// given character is a vowel 
bool isVowel(char ch) 
    if (ch == 'a' or ch == 'e'
        or ch == 'i' or ch == 'o'
        or ch == 'u'
        return true
    else
        return false
  
// Function to return the minimum 
// cost to convert all the vowels 
// of a string to a single one 
int minCost(string S) 
    // Stores count of 
    // respective vowels 
    int cA = 0; 
    int cE = 0; 
    int cI = 0; 
    int cO = 0; 
    int cU = 0; 
  
    // Iterate through the string 
    for (int i = 0; i < S.size(); i++) { 
  
        // If a vowel is encountered 
        if (isVowel(S[i])) { 
  
            // Calculate the cost 
            cA += abs(S[i] - 'a'); 
            cE += abs(S[i] - 'e'); 
            cI += abs(S[i] - 'i'); 
            cO += abs(S[i] - 'o'); 
            cU += abs(S[i] - 'u'); 
        
    
  
    // Return the minimum cost 
    return min(min(min(min(cA, cE), 
                    cI), 
                cO), 
            cU); 
  
// Driver Code 
int main() 
    string S = "geeksforgeeks"
  
    cout << minCost(S) << endl; 
  
    return 0; 

Java




// Java program for the above approach 
import java.util.*;
  
class GFG{
  
// Function that return true if the
// given character is a vowel
static boolean isVowel(char ch)
{
    if (ch == 'a' || ch == 'e' || 
        ch == 'i' || ch == 'o' ||
        ch == 'u')
        return true;
    else
        return false;
}
  
// Function to return the minimum
// cost to convert all the vowels
// of a string to a single one
static int minCost(String S)
{
      
    // Stores count of
    // respective vowels
    int cA = 0;
    int cE = 0;
    int cI = 0;
    int cO = 0;
    int cU = 0;
  
    // Iterate through the string
    for(int i = 0; i < S.length(); i++)
    {
          
        // If a vowel is encountered
        if (isVowel(S.charAt(i)))
        {
              
            // Calculate the cost
            cA += Math.abs(S.charAt(i) - 'a');
            cE += Math.abs(S.charAt(i) - 'e');
            cI += Math.abs(S.charAt(i) - 'i');
            cO += Math.abs(S.charAt(i) - 'o');
            cU += Math.abs(S.charAt(i) - 'u');
        }
    }
  
    // Return the minimum cost
    return Math.min(Math.min(Math.min(Math.min(cA, cE), 
                                         cI), cO), cU);
}
  
// Driver code
public static void main(String[] args)
{
    String S = "geeksforgeeks";
  
    System.out.println(minCost(S));
}
}
  
// This code is contributed by offbeat

Python3




# Python3 program for the above approach 
  
# Function that return true if the 
# given character is a vowel 
def isVowel(ch):
  
    if (ch == 'a' or ch == 'e' or
        ch == 'i' or ch == 'o' or 
        ch == 'u'):
        return True
    else:
        return False
  
# Function to return the minimum 
# cost to convert all the vowels 
# of a string to a single one 
def minCost(S):
  
    # Stores count of 
    # respective vowels 
    cA = 0
    cE = 0
    cI = 0
    cO = 0
    cU = 0
  
    # Iterate through the string 
    for i in range(len(S)): 
  
        # If a vowel is encountered 
        if (isVowel(S[i])): 
  
            # Calculate the cost 
            cA += abs(ord(S[i]) - ord('a')); 
            cE += abs(ord(S[i]) - ord('e')); 
            cI += abs(ord(S[i]) - ord('i')); 
            cO += abs(ord(S[i]) - ord('o')); 
            cU += abs(ord(S[i]) - ord('u')); 
          
    # Return the minimum cost 
    return min(min(min(min(cA, cE), cI), cO), cU); 
  
# Driver code
S = "geeksforgeeks"
  
print(minCost(S)) 
  
# This code is contributed by grand_master

C#




// C# program for the above approach 
using System;
  
class GFG{
  
// Function that return true if the
// given character is a vowel
static bool isVowel(char ch)
{
    if (ch == 'a' || ch == 'e' || 
        ch == 'i' || ch == 'o' ||
        ch == 'u')
        return true;
    else
        return false;
}
  
// Function to return the minimum
// cost to convert all the vowels
// of a string to a single one
static int minCost(String S)
{
      
    // Stores count of
    // respective vowels
    int cA = 0;
    int cE = 0;
    int cI = 0;
    int cO = 0;
    int cU = 0;
  
    // Iterate through the string
    for(int i = 0; i < S.Length; i++)
    {
          
        // If a vowel is encountered
        if (isVowel(S[i]))
        {
              
            // Calculate the cost
            cA += Math.Abs(S[i] - 'a');
            cE += Math.Abs(S[i] - 'e');
            cI += Math.Abs(S[i] - 'i');
            cO += Math.Abs(S[i] - 'o');
            cU += Math.Abs(S[i] - 'u');
        }
    }
  
    // Return the minimum cost
    return Math.Min(Math.Min(
           Math.Min(Math.Min(cA, cE), 
                       cI), cO), cU);
}
  
// Driver code
public static void Main(String[] args)
{
    String S = "geeksforgeeks";
  
    Console.WriteLine(minCost(S));
}
}
  
// This code is contributed by Rajput-Ji
Output: 
10

 

Time Complexity: O(N) 
Auxiliary Space: O(1)
 

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