Minimize cost to replace all the vowels of a given String by a single vowel

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: 6
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# 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

Javascript

<script>
 
// Javascript program f|| the above approach 
 
// Function that return true if the 
// given character is a vowel 

function isVowel(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 

function minCost(S) 
{ 

    // St||es count of 

    // respective vowels 

    var cA = 0; 

    var cE = 0; 

    var cI = 0; 

    var cO = 0; 

    var cU = 0; 
 
    // Iterate through the string 

    for(var i = 0; i < S.length; i++) { 
 
        // If a vowel is encountered 

        if (isVowel(S[i])) { 
 
            // Calculate the cost 

            cA += Math.abs(S.charCodeAt(i) - 'a'.charCodeAt(0)); 

            cE += Math.abs(S.charCodeAt(i) - 'e'.charCodeAt(0)); 

            cI += Math.abs(S.charCodeAt(i) - 'i'.charCodeAt(0)); 

            cO += Math.abs(S.charCodeAt(i) - 'o'.charCodeAt(0)); 

            cU += Math.abs(S.charCodeAt(i) - 'u'.charCodeAt(0)); 

        } 

    } 
 
    // Return the Math.minimum cost 

    return Math.min(Math.min(Math.min(Math.min(cA, cE), 

                    cI), 

                cO), 

            cU); 
} 
 
// Driver Code 

var S = "geeksforgeeks"; 
document.write(minCost(S)); 
 
// This code is contributed by importantly.
</script>

Output:

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

Article Tags :

DSA

Strings

frequency-counting