# Minimize changes to make all characters equal by changing vowel to consonant and vice versa

Given a string str[] of lower-case characters, the task is to make all characters of the string equal in the minimum number of operations such that in each operation either choose a vowel and change it to a consonant or vice-versa.

Examples:

Input: str[] = “geeksforgeeks”
Output: 10
Explanation: To make all the characters equal, make the following changes –

1. Change ‘o’ to a consonant(let’s say) ‘z’ and then to ‘e’
2. Change every other consonant(‘g’, ‘k’, ‘s’, ‘ f’, ‘r’, ) to ‘e’

This results in the string str = “eeeeeeeeeeeee” and the total number of operations performed is 10.

Input: str[] = “coding”
Output: 6

Approach: It can be deduced from the problem statement that in order to change a consonant to a vowel or vice versa, 1 operation is required. In order to change a vowel to a vowel or a consonant to a consonant, 2 operations are required. Because for changing vowel to vowel, first we need to change vowel -> consonant and then consonant -> vowel. Similarly, for changing consonant to consonant, first we need to change consonant -> vowel and then vowel -> consonant . The idea would be to maintain two variables in parallel :-

• Cv = the cost of changing all the characters to the maximum occurring vowel = no. of consonants + ( total number of vowels – frequency of maximum occurring vowel  ) * 2
• Cc = the cost of changing all the characters to the maximum occurring consonant = no. of vowels + (total number of consonants – frequency of maximum occurring consonant) * 2

Now the minimum of these 2 i.e., min(Cv, Cc) will give the required minimum number of steps in which we can transform the string. Follow the steps below to solve the problem:

• Initialize the variable ans, vowels and consonants as 0 to store the answer, number of vowels and the number of consonants.
• Initialize 2 variables max_vowels and max_consonants as INT_MIN to find the maximum occurring vowel and maximum occurring consonant in the given string.
• Initialize 2 unordered_map<char, int> freq_vowels[] and freq_consonant[] to store the frequency of vowels and consonants.
• Iterate over the range [0, N) using the variable i and perform the following steps:
• If the current character is a vowel, then increase the count of vowels by 1 and it’s frequency in the map by 1 otherwise do it for the consonant.
• Traverse both the unordered_maps and find the maximum occurring vowel and consonant.
• Using the above formula, calculate the ans.
• After performing the above steps, print the value of ans as the answer.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach   #include using namespace std;   // Function to find the minimum number // of steps to make all characters equal int operations(string s) {       // Initializing the variables     int ans = 0;     int vowels = 0, consonants = 0;     int max_vowels = INT_MIN;     int max_consonants = INT_MIN;       // Store the frequency     unordered_map freq_consonants;     unordered_map freq_vowels;       // Iterate over the string     for (int i = 0; i < s.size(); i++) {         if (s[i] == 'a' or s[i] == 'e' or s[i] == 'i'             or s[i] == 'o' or s[i] == 'u') {               // Calculate the total             // number of vowels             vowels += 1;               // Storing frequency of             // each vowel             freq_vowels[s[i]] += 1;         }         else {               // Count the consonants             consonants += 1;               // Storing the frequency of             // each consonant             freq_consonants[s[i]] += 1;         }     }       // Iterate over the 2 maps     for (auto it = freq_consonants.begin();          it != freq_consonants.end(); it++) {           // Maximum frequency of consonant         max_consonants = max(             max_consonants,             it->second);     }     for (auto it = freq_vowels.begin();          it != freq_vowels.end(); it++) {           // Maximum frequency of vowel         max_vowels             = max(max_vowels,                   it->second);     }       // Find the result     ans = min((2 * (vowels - max_vowels)                + consonants),               (2 * (consonants - max_vowels)                + consonants));       return ans; }   // Driver Code int main() {     string S = "geeksforgeeks";     cout << operations(S);       return 0; }

## Java

 // Java program to implement the above approach import java.io.*; import java.util.*;   class GFG {     // Function to find the minimum number of steps to make   // all characters equal   static int operations(String s)   {       // Initializign the variables     int ans = 0;     int vowels = 0, consonants = 0;     int max_vowels = Integer.MIN_VALUE;     int max_consonants = Integer.MIN_VALUE;       // Store the frequency     HashMap freq_consonants       = new HashMap<>();     HashMap freq_vowels       = new HashMap<>();       // Iterate over the string     for (int i = 0; i < s.length(); i++) {       if (s.charAt(i) == 'a' || s.charAt(i) == 'e'           || s.charAt(i) == 'i' || s.charAt(i) == 'o'           || s.charAt(i) == 'u') {           // Calculate the total         // number of vowels         vowels += 1;           // Storing frequency of         // each vowel         freq_vowels.put(           s.charAt(i),           freq_vowels.getOrDefault(s.charAt(i), 0)           + 1);       }       else {           // Count the consonants         consonants += 1;           // Storing the frequency of         // each consonant         freq_consonants.put(           s.charAt(i),           freq_consonants.getOrDefault(             s.charAt(i), 0)           + 1);       }     }       // Iterate over the 2 maps     for (Map.Entry it :          freq_consonants.entrySet()) {         // Maximum frequency of consonant       max_consonants         = Math.max(max_consonants, it.getValue());     }       for (Map.Entry it :          freq_vowels.entrySet()) {         // Maximum frequency of vowel       max_vowels         = Math.max(max_vowels, it.getValue());     }       // Find the result     ans = Math.min(       (2 * (vowels - max_vowels) + consonants),       (2 * (consonants - max_vowels) + consonants));       return ans;   }     public static void main(String[] args)   {     String S = "geeksforgeeks";     System.out.print(operations(S));   } }   // This code is contributed by lokeshmvs21.

## Python3

 # Python 3 program for the above approach import sys from collections import defaultdict   # Function to find the minimum number # of steps to make all characters equal def operations(s):       # Initializing the variables     ans = 0     vowels = 0     consonants = 0     max_vowels = -sys.maxsize-1     max_consonants = -sys.maxsize-1       # Store the frequency     freq_consonants = defaultdict(int)     freq_vowels = defaultdict(int)       # Iterate over the string     for i in range(len(s)):         if (s[i] == 'a' or s[i] == 'e' or s[i] == 'i'                 or s[i] == 'o' or s[i] == 'u'):               # Calculate the total             # number of vowels             vowels += 1               # Storing frequency of             # each vowel             freq_vowels[s[i]] += 1           else:               # Count the consonants             consonants += 1               # Storing the frequency of             # each consonant             freq_consonants[s[i]] += 1       # Iterate over the 2 maps     for it in freq_consonants:           # Maximum frequency of consonant         max_consonants = max(             max_consonants,             freq_consonants[it])       for it in freq_vowels:           # Maximum frequency of vowel         max_vowels = max(max_vowels,                          freq_vowels[it])       # Find the result     ans = min((2 * (vowels - max_vowels)                + consonants),               (2 * (consonants - max_vowels)                + consonants))       return ans   # Driver Code if __name__ == "__main__":       S = "geeksforgeeks"     print(operations(S))       # This code is contributed by ukasp.



## C#

 using System; using System.Collections.Generic;   class Program {     // Function to find the minimum number     // of steps to make all characters equal     static int Operations(string s)     {         // Initializing the variables         int ans = 0;         int vowels = 0, consonants = 0;         int maxVowels = int.MinValue;         int maxConsonants = int.MinValue;           // Store the frequency         var freqConsonants = new Dictionary();         var freqVowels = new Dictionary();           // Iterate over the string         for (int i = 0; i < s.Length; i++) {             if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i'                 || s[i] == 'o' || s[i] == 'u') {                 // Calculate the total                 // number of vowels                 vowels += 1;                   // Storing frequency of                 // each vowel                 if (!freqVowels.ContainsKey(s[i])) {                     freqVowels[s[i]] = 1;                 }                 else {                     freqVowels[s[i]] += 1;                 }             }             else {                 // Count the consonants                 consonants += 1;                   // Storing the frequency of                 // each consonant                 if (!freqConsonants.ContainsKey(s[i])) {                     freqConsonants[s[i]] = 1;                 }                 else {                     freqConsonants[s[i]] += 1;                 }             }         }           // Iterate over the 2 dictionaries         foreach(var pair in freqConsonants)         {             // Maximum frequency of consonant             maxConsonants                 = Math.Max(maxConsonants, pair.Value);         }         foreach(var pair in freqVowels)         {             // Maximum frequency of vowel             maxVowels = Math.Max(maxVowels, pair.Value);         }           // Find the result         ans = Math.Min(             (2 * (vowels - maxVowels) + consonants),             (2 * (consonants - maxVowels) + consonants));           return ans;     }       // Driver Code     static void Main()     {         string S = "geeksforgeeks";         Console.WriteLine(Operations(S));     } }

Output:

10

Time Complexity: O(N)
Auxiliary Space: O(1) as it is using constant space for variables and map to store the frequency of characters

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next