# Minimum characters to be replaced to make frequency of all characters same

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”

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 = { 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``; ` `    ``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 `

Output:
```1
```

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