Make all characters in given string equal by changing vowel into consonant and vice versa
Given a string str containing lowercase characters, the task is to find the minimum number of operations needed to make all the characters equal. In one operation, any consonant can be converted to any vowel or any vowel can be converted to any consonant.
Examples:
Input: str = “banana”
Output: 3
Explanation: Convert all consonants to Vowel character AInput: str = “hexon”
Output: 5
Explanation: Convert E to O first then all the consonants to Alphabet O
Approach: The basic catch here is that:
- The operations required to change a consonant to another consonant or a vowel to another vowel: 2 Operation
- The operation required to change a consonant to a vowel or a vowel to consonant: 1 Operation
So, it’s clear that it is more economical to change a consonant to a vowel or a vowel to a consonant than changing a consonant to another consonant or a vowel to another vowel.
Now to solve this question, follow the steps below:
- Calculate the frequency of all the characters and find the consonant and vowels with the highest frequency, let’s say A and B respectively.
- Try to change all the characters to A and B, and store the operations required in variables minA and minB respectively.
- Minimum of minA and minB is the answer to this problem.
Below is the implementation of the above approach.
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to return the minimum operations // required to make all string characters equal int minOperations(string str) { int n = str.size(); // Vector to store the // frequency of all characters vector< int > freq(26, 0); for ( auto x : str) { freq[x - 'a' ]++; } int mxA = INT_MIN, mxB = INT_MIN; // Variables to store // consonant and vowel // with highest frequency char A, B; vector< char > vowels = { 'a' , 'e' , 'i' , 'o' , 'u' }; for ( int i = 0; i < 26; ++i) { bool isVowel = 0; for ( auto x : vowels) { if ( 'a' + i == x) { isVowel = 1; break ; } } // If current character is a vowel if (isVowel) { if (mxB < freq[i]) { mxB = freq[i]; B = 'a' + i; } } // If current character is a consonant else { if (mxA < freq[i]) { mxA = freq[i]; A = 'a' + i; } } } int minA = 0, minB = 0; for ( auto x : str) { bool isVowel = 0; for ( auto y : vowels) { if (x == y) { isVowel = 1; break ; } } // If current character is a vowel if (isVowel) { if (x != B) { minB += 2; } minA += 1; } // If current character is a // consonant else { if (x != A) { minA += 2; } minB += 1; } } // If no vowel exists if (mxB == INT_MIN) { return minA; } // If no consonant exists if (mxA == INT_MIN) { return minB; } // Returning minimum of the two return min(minA, minB); } // Driver Code int main() { string str = "hexon" ; cout << minOperations(str); } |
Java
// Java program for the above approach import java.util.*; public class GFG { // Function to return the minimum operations // required to make all string characters equal static int minOperations(String str) { int n = str.length(); // Vector to store the // frequency of all characters char [] freq = new char [ 26 ]; for ( char x : str.toCharArray()) { freq[x - 'a' ]++; } int mxA = Integer.MIN_VALUE, mxB = Integer.MIN_VALUE; // Variables to store // consonant and vowel // with highest frequency char A = ' ' , B = ' ' ; char [] vowels = { 'a' , 'e' , 'i' , 'o' , 'u' }; for ( int i = 0 ; i < 26 ; ++i) { int isVowel = 0 ; for ( char x : vowels) { if ( 'a' + i == x) { isVowel = 1 ; break ; } } // If current character is a vowel if (isVowel > 0 ) { if (mxB < freq[i]) { mxB = freq[i]; B = ( char )( 97 + i); } } // If current character is a consonant else { if (mxA < freq[i]) { mxA = freq[i]; A = ( char )( 97 + i); } } } int minA = 0 , minB = 0 ; for ( char x : str.toCharArray()) { int isVowel = 0 ; for ( char y : vowels) { if (x == y) { isVowel = 1 ; break ; } } // If current character is a vowel if (isVowel > 0 ) { if (x != B) { minB += 2 ; } minA += 1 ; } // If current character is a // consonant else { if (x != A) { minA += 2 ; } minB += 1 ; } } // If no vowel exists if (mxB == Integer.MIN_VALUE) { return minA; } // If no consonant exists if (mxA == Integer.MIN_VALUE) { return minB; } // Returning minimum of the two return Math.min(minA, minB); } // Driver Code public static void main(String args[]) { String str = "hexon" ; System.out.println(minOperations(str)); } } // This code is contributed by Samim Hossain Mondal. |
Python3
# python program for the above approach INT_MIN = - 2147483647 - 1 # Function to return the minimum operations # required to make all string characters equal def minOperations( str ): n = len ( str ) # Vector to store the # frequency of all characters freq = [ 0 for _ in range ( 26 )] for x in str : freq[ ord (x) - ord ( 'a' )] + = 1 mxA = INT_MIN mxB = INT_MIN # Variables to store # consonant and vowel # with highest frequency A = '' B = '' vowels = [ 'a' , 'e' , 'i' , 'o' , 'u' ] for i in range ( 0 , 26 ): isVowel = 0 for x in vowels: if ( ord ( 'a' ) + i = = ord (x)): isVowel = 1 break # If current character is a vowel if (isVowel): if (mxB < freq[i]): mxB = freq[i] B = chr ( ord ( 'a' ) + i) # If current character is a consonant else : if (mxA < freq[i]): mxA = freq[i] A = chr ( ord ( 'a' ) + i) minA = 0 minB = 0 for x in str : isVowel = 0 for y in vowels: if (x = = y): isVowel = 1 break # If current character is a vowel if (isVowel): if (x ! = B): minB + = 2 minA + = 1 # If current character is a # consonant else : if (x ! = A): minA + = 2 minB + = 1 # If no vowel exists if (mxB = = INT_MIN): return minA # If no consonant exists if (mxA = = INT_MIN): return minB # Returning minimum of the two return min (minA, minB) # Driver Code if __name__ = = "__main__" : str = "hexon" print (minOperations( str )) # This code is contributed by rakeshsahni |
C#
// C# program for the above approach using System; class GFG { // Function to return the minimum operations // required to make all string characters equal static int minOperations( string str) { int n = str.Length; // Vector to store the // frequency of all characters char [] freq = new char [26]; foreach ( char x in str) { freq[x - 'a' ]++; } int mxA = Int32.MinValue, mxB = Int32.MinValue; // Variables to store // consonant and vowel // with highest frequency char A = ' ' , B = ' ' ; char [] vowels = { 'a' , 'e' , 'i' , 'o' , 'u' }; for ( int i = 0; i < 26; ++i) { int isVowel = 0; foreach ( char x in vowels) { if ( 'a' + i == x) { isVowel = 1; break ; } } // If current character is a vowel if (isVowel > 0) { if (mxB < freq[i]) { mxB = freq[i]; B = ( char )(97 + i); } } // If current character is a consonant else { if (mxA < freq[i]) { mxA = freq[i]; A = ( char )(97 + i); } } } int minA = 0, minB = 0; foreach ( char x in str) { int isVowel = 0; foreach ( char y in vowels) { if (x == y) { isVowel = 1; break ; } } // If current character is a vowel if (isVowel > 0) { if (x != B) { minB += 2; } minA += 1; } // If current character is a // consonant else { if (x != A) { minA += 2; } minB += 1; } } // If no vowel exists if (mxB == Int32.MinValue) { return minA; } // If no consonant exists if (mxA == Int32.MinValue) { return minB; } // Returning minimum of the two return Math.Min(minA, minB); } // Driver Code public static void Main() { string str = "hexon" ; Console.WriteLine(minOperations(str)); } } // This code is contributed by ukasp. |
Javascript
<script> // Javascript program for the above approach // Function to return the minimum operations // required to make all string characters equal function minOperations(str) { let n = str.length; // Vector to store the // frequency of all characters let freq = new Array(26).fill(0); for (x of str) { freq[x.charCodeAt(0) - 'a' .charCodeAt(0)]++; } let mxA = Number.MIN_SAFE_INTEGER, mxB = Number.MIN_SAFE_INTEGER; // Variables to store // consonant and vowel // with highest frequency let A = "" , B = "" ; let vowels = [ 'a' , 'e' , 'i' , 'o' , 'u' ]; for (let i = 0; i < 26; ++i) { let isVowel = 0; for (x of vowels) { if ( 'a' .charCodeAt(0) + i == x.charCodeAt(0)) { isVowel = 1; break ; } } // If current character is a vowel if (isVowel) { if (mxB < freq[i]) { mxB = freq[i]; B = String.fromCharCode( 'a' .charCodeAt(0) + i); } } // If current character is a consonant else { if (mxA < freq[i]) { mxA = freq[i]; A = String.fromCharCode( 'a' .charCodeAt(0) + i); } } } let minA = 0, minB = 0; for (x of str) { let isVowel = 0; for (y of vowels) { if (x == y) { isVowel = 1; break ; } } // If current character is a vowel if (isVowel) { if (x != B) { minB += 2; } minA += 1; } // If current character is a // consonant else { if (x != A) { minA += 2; } minB += 1; } } // If no vowel exists if (mxB == Number.MIN_SAFE_INTEGER) { return minA; } // If no consonant exists if (mxA == Number.MIN_SAFE_INTEGER) { return minB; } // Returning minimum of the two return Math.min(minA, minB); } // Driver Code let str = "hexon" ; document.write(minOperations(str)) // This code is contributed by saurabh_jaiswal. </script> |
5
Time Complexity: O(N)
Auxiliary Space: O(1)
Another Approach:
- Iterate over each character in the input string and count the number of vowels and consonants separately.
- If either the vowel count or consonant count is zero, return 0 because all characters are already equal.
- If the vowel count and consonant count are equal, return either count since any character can be converted to the other.
- Otherwise, return the minimum of the vowel count and consonant count multiplied by 2, plus 1 to account for the remaining characters that need to be converted to the opposite type.
- This minimum count represents the minimum number of operations needed to make all characters in the string equal.
Below is the implementation of the above approach:
C++
#include <iostream> #include <string> #include <algorithm> bool isVowel( char c) { c = tolower (c); return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' ); } int minOperationsToMakeAllCharactersEqual( const std::string& str) { int vowelCount = 0; int consonantCount = 0; for ( char c : str) { if (isVowel(c)) { vowelCount++; } else { consonantCount++; } } if (vowelCount == 0 || consonantCount == 0) { return 0; // All characters are already equal } if (vowelCount == consonantCount) { return vowelCount; // Both vowels and consonants are equal, any character can be converted to the other } return std::min(vowelCount, consonantCount) * 2 + 1; } int main() { std::string str = "hexon" ; int minOperations = minOperationsToMakeAllCharactersEqual(str); std::cout << minOperations << std::endl; return 0; } // This code is contributed by rudra1807raj |
5
Time Complexity: O(N).
Auxiliary Space: O(1).
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