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

• Last Updated : 19 May, 2021

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 using namespace std; // Function that return true if the// given character is a vowelbool 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 oneint 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 Codeint main(){    string S = "geeksforgeeks";     cout << minCost(S) << endl;     return 0;}

Java

 // Java program for the above approachimport java.util.*; class GFG{ // Function that return true if the// given character is a vowelstatic 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 onestatic 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 codepublic 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 voweldef 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 onedef 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 codeS = "geeksforgeeks"; print(minCost(S)) # This code is contributed by grand_master

C#

 // C# program for the above approachusing System; class GFG{ // Function that return true if the// given character is a vowelstatic 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 onestatic 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 codepublic static void Main(String[] args){    String S = "geeksforgeeks";     Console.WriteLine(minCost(S));}} // This code is contributed by Rajput-Ji

Javascript


Output:
10

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

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