Given a string S consisting of alphabets [‘A’ – ‘Z’], the task is to find the minimum number of operations required to make frequency of every character equal. In one operation any character of the string can be chosen and replaced with another valid character.
Examples:
Input: S = “ABCB”
Output: 1
Explanation:
In the given string character ‘C’ can be replaced by ‘A’, such that occurrence of every character becomes equal to 2.
Updated String = “ABAB”
Input: S = “BBC”
Output : 1
Explanation:
In the given string character ‘C’ can be replaced by ‘B’, such that occurrence of every character becomes equal to 3.
Updated string = “BBB”
Approach: The idea is to find the frequency of every character in the string and then sort the characters according to their frequencies in descending order. Finally, we can check for each character of string which yields the minimum number of characters to be changed and print this minimum number of characters to be changed.
Below is the implementation of the above approach:
C++
// C++ implementation to find the// Minimum characters to be replaced// to make frequency of all characters same#include <bits/stdc++.h>using namespace std;// Function to find the// Minimum operations to convert// given string to another with// equal frequencies of charactersint minOperations(string s){ // Frequency of characters int freq[26] = { 0 }; int n = s.length(); // Loop to find the Frequency // of each character for (int i = 0; i < n; i++) { freq[s[i] - 'A']++; } // Sort in decreasing order // based on frequency sort(freq, freq + 26, greater<int>()); // Maximum possible answer int answer = n; // Loop to find the minimum operations // required such that frequency of // every character is equal for (int i = 1; i <= 26; i++) { if (n % i == 0) { int x = n / i; int y = 0; for (int j = 0; j < i; j++) { y += min(freq[j], x); } answer = min(answer, n - y); } } return answer;}// Driver Codeint main(){ string s = "BBC"; cout << minOperations(s); return 0;} |
Java
// Java implementation to find the// Minimum characters to be replaced// to make frequency of all characters same import java.util.*;class GFG{ // Function to find the// Minimum operations to convert// given String to another with// equal frequencies of charactersstatic int minOperations(String s){ // Frequency of characters Integer freq[] = new Integer[26]; Arrays.fill(freq, 0); int n = s.length(); // Loop to find the Frequency // of each character for (int i = 0; i < n; i++) { freq[s.charAt(i) - 'A']++; } // Sort in decreasing order // based on frequency Arrays.sort(freq, Collections.reverseOrder()); // Maximum possible answer int answer = n; // Loop to find the minimum operations // required such that frequency of // every character is equal for (int i = 1; i <= 26; i++) { if (n % i == 0) { int x = n / i; int y = 0; for (int j = 0; j < i; j++) { y += Math.min(freq[j], x); } answer = Math.min(answer, n - y); } } return answer;} // Driver Codepublic static void main(String[] args){ String s = "BBC"; System.out.print(minOperations(s)); }}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation to find the# minimum characters to be replaced# to make frequency of all characters same# Function to find the minimum # operations to convert given # string to another with equal # frequencies of charactersdef minOperations(s): # Frequency of characters freq = [0] * 26 n = len(s) # Loop to find the Frequency # of each character for i in range(n): freq[ord(s[i]) - ord('A')] += 1 # Sort in decreasing order # based on frequency freq.sort(reverse = True) # Maximum possible answer answer = n # Loop to find the minimum operations # required such that frequency of # every character is equal for i in range(1, 27): if (n % i == 0): x = n // i y = 0 for j in range(i): y += min(freq[j], x) answer = min(answer, n - y) return answer# Driver Codeif __name__ == "__main__": s = "BBC" print (minOperations(s))# This code is contributed by chitranayal |
C#
// C# implementation to find the minimum// characters to be replaced to make // frequency of all characters sameusing System;class GFG{// Function to find the minimum// operations to convert given// string to another with equal// frequencies of charactersstatic int minOperations(String s){ // Frequency of characters int []freq = new int[26]; int n = s.Length; // Loop to find the frequency // of each character for(int i = 0; i < n; i++) { freq[s[i] - 'A']++; } // Sort in decreasing order // based on frequency Array.Sort(freq); Array.Reverse(freq); // Maximum possible answer int answer = n; // Loop to find the minimum operations // required such that frequency of // every character is equal for(int i = 1; i <= 26; i++) { if (n % i == 0) { int x = n / i; int y = 0; for(int j = 0; j < i; j++) { y += Math.Min(freq[j], x); } answer = Math.Min(answer, n - y); } } return answer;}// Driver Codepublic static void Main(String[] args){ String s = "BBC"; Console.Write(minOperations(s));}}// This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript implementation to find the minimum // characters to be replaced to make // frequency of all characters same // Function to find the minimum // operations to convert given // string to another with equal // frequencies of characters function minOperations(s) { // Frequency of characters var freq = new Array(26).fill(0); var n = s.length; // Loop to find the frequency // of each character for (var i = 0; i < n; i++) { freq[s[i].charCodeAt(0) - "A".charCodeAt(0)]++; } // Sort in decreasing order // based on frequency freq.sort((a, b) => b - a); // Maximum possible answer var answer = n; // Loop to find the minimum operations // required such that frequency of // every character is equal for (var i = 1; i <= 26; i++) { if (n % i === 0) { var x = n / i; var y = 0; for (var j = 0; j < i; j++) { y += Math.min(freq[j], x); } answer = Math.min(answer, n - y); } } return answer; } // Driver Code var s = "BBC"; document.write(minOperations(s)); </script> |
Output:
1
Time Complexity: O(n)
Auxiliary Space: O(26)
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