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++ 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 characters int 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 Code int main()
{ string s = "BBC" ;
cout << minOperations(s);
return 0;
} |
// 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 characters static 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 Code public static void main(String[] args)
{ String s = "BBC" ;
System.out.print(minOperations(s));
} } // This code is contributed by 29AjayKumar |
# 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 characters def 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 Code if __name__ = = "__main__" :
s = "BBC"
print (minOperations(s))
# This code is contributed by chitranayal |
// C# implementation to find the minimum // characters to be replaced to make // frequency of all characters same using System;
class GFG{
// Function to find the minimum // operations to convert given // string to another with equal // frequencies of characters static 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 Code public static void Main(String[] args)
{ String s = "BBC" ;
Console.Write(minOperations(s));
} } // This code is contributed by Rajput-Ji |
<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> |
1
Time Complexity: O(n)
Auxiliary Space: O(26)