# Minimum operations to make frequency of all characters equal K

Given a string S of length N. The task is to find the minimum number of steps required on strings, so that it has exactly K different alphabets all with the same frequency.

Note: In one step, we can change a letter to any other letter.

Examples:

```Input: S = "abbc", N = 4, K = 2
Output: 1
In one step convert 'c' to 'a'. Hence string has
two different letters a and b both
occurring 2 times.
```

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

Approach:

1. Check if K divides N, then only it is possible to convert the given string, otherwise not.
2. Maintain the count of all alphabets present in string S, in an array A.
3. Evaluate E = N/K, the frequency with which alphabets will be present in the final string.
4. Separate the alphabets with frequency more than or equal to E and less than E in two parts.
5. Maintain the number of steps required for each alphabet to convert its count to E, sort these vector obtained in above step.
6. Lastly, take all possibility to pick:
```Set 1 : 0   Set 2 : K
Set 1 : 1   Set 2 : K-1 .... so on
```
7. Keep a ans variable to calculate minimum number of steps among all possibility in step 6.
8. Say L1 is the number of operation required on Set 1, L2 is the number of operations required on set 2. Then total operations required is maximum of L1, L2 . As suppose ‘a’ is required one less in string while ‘b’ is required one more than we can change ‘a’ to ‘b’, thus reducing number of steps.

Below is the implementation of the above approach:

## C++

 `// C++ program to convert the given string ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the minimum number of ` `// operations to convert the given string ` `void` `minOperation(string S, ``int` `N, ``int` `K) ` `{ ` `    ``// Check if N is divisible by K ` `    ``if` `(N % K) { ` `        ``cout << ``"Not Possible"` `<< endl; ` `        ``return``; ` `    ``} ` ` `  `    ``// Array to store frequency of characters ` `    ``// in given string ` `    ``int` `count[26] = { 0 }; ` `    ``for` `(``int` `i = 0; i < N; i++) { ` `        ``count[S[i] - 97]++; ` `    ``} ` ` `  `    ``int` `E = N / K; ` ` `  `    ``vector<``int``> greaterE; ` `    ``vector<``int``> lessE; ` ` `  `    ``for` `(``int` `i = 0; i < 26; i++) { ` ` `  `        ``// Two arrays with number of operations ` `        ``// required ` `        ``if` `(count[i] < E) ` `            ``lessE.push_back(E - count[i]); ` `        ``else` `            ``greaterE.push_back(count[i] - E); ` `    ``} ` ` `  `    ``sort(greaterE.begin(), greaterE.end()); ` `    ``sort(lessE.begin(), lessE.end()); ` ` `  `    ``int` `mi = INT_MAX; ` ` `  `    ``for` `(``int` `i = 0; i <= K; i++) { ` ` `  `        ``// Checking for all possibility ` `        ``int` `set1 = i; ` `        ``int` `set2 = K - i; ` ` `  `        ``if` `(greaterE.size() >= set1 && lessE.size() >= set2) { ` ` `  `            ``int` `step1 = 0; ` `            ``int` `step2 = 0; ` ` `  `            ``for` `(``int` `j = 0; j < set1; j++) ` `                ``step1 += greaterE[j]; ` ` `  `            ``for` `(``int` `j = 0; j < set2; j++) ` `                ``step2 += lessE[j]; ` ` `  `            ``mi = min(mi, max(step1, step2)); ` `        ``} ` `    ``} ` ` `  `    ``cout << mi << endl; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string S = ``"accb"``; ` `    ``int` `N = S.size(); ` `    ``int` `K = 2; ` ` `  `    ``minOperation(S, N, K); ` ` `  `    ``return` `0; ` `} `

## Java

 `// JAVA program to convert the given string ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to find the minimum number of ` `    ``// operations to convert the given string ` `    ``static` `void` `minOperation(String S, ``int` `N, ``int` `K) ` `    ``{ ` `        ``// Check if N is divisible by K ` `        ``if` `(N % K != ``0``) ` `        ``{ ` `            ``System.out.println(``"Not Possible"``); ` `        ``} ` `        ``else` `        ``{ ` `            ``// Array to store frequency of characters ` `            ``// in given string ` `            ``int` `[] count = ``new` `int``[``26``]; ` `             `  `            ``for` `(``int` `i = ``0``; i < N; i++) ` `            ``{ ` `                ``count[(S.charAt(i) - ``97``)]++; ` `            ``} ` `         `  `            ``int` `E = N / K; ` `         `  `            ``Vector greaterE = ``new` `Vector<>();  ` `            ``Vector lessE = ``new` `Vector<>();  ` `         `  `            ``for` `(``int` `i = ``0``; i < ``26``; i++)  ` `            ``{ ` `         `  `                ``// Two arrays with number of operations ` `                ``// required ` `                ``if` `(count[i] < E) ` `                    ``lessE.add(E - count[i]); ` `                ``else` `                    ``greaterE.add(count[i] - E); ` `            ``} ` `         `  `            ``Collections.sort(greaterE); ` `            ``Collections.sort(lessE); ` `         `  `            ``int` `mi = Integer.MAX_VALUE; ` `         `  `            ``for` `(``int` `i = ``0``; i <= K; i++) ` `            ``{ ` `         `  `                ``// Checking for all possibility ` `                ``int` `set1 = i; ` `                ``int` `set2 = K - i; ` `         `  `                ``if` `(greaterE.size() >= set1 &&  ` `                            ``lessE.size() >= set2)  ` `                ``{ ` `         `  `                    ``int` `step1 = ``0``; ` `                    ``int` `step2 = ``0``; ` `         `  `                    ``for` `(``int` `j = ``0``; j < set1; j++) ` `                        ``step1 += greaterE.get(j); ` `         `  `                    ``for` `(``int` `j = ``0``; j < set2; j++) ` `                        ``step2 += lessE.get(j); ` `         `  `                    ``mi = Math.min(mi, Math.max(step1, step2)); ` `                ``} ` `            ``} ` `         `  `            ``System.out.println(mi); ` `        ``} ` ` `  `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``String S = ``"accb"``; ` `        ``int` `N = S.length(); ` `        ``int` `K = ``2``; ` `     `  `        ``minOperation(S, N, K); ` `    ``} ` `} ` ` `  `// This code is contributed by ihritik `

## Python3

 `# Python3 program to convert the given string  ` ` `  `# Function to find the minimum number of  ` `# operations to convert the given string  ` `def` `minOperation(S, N, K):  ` ` `  `    ``# Check if N is divisible by K  ` `    ``if` `N ``%` `K:  ` `        ``print``(``"Not Possible"``)  ` `        ``return` ` `  `    ``# Array to store frequency of  ` `    ``# characters in given string  ` `    ``count ``=` `[``0``] ``*` `26` `    ``for` `i ``in` `range``(``0``, N):  ` `        ``count[``ord``(S[i]) ``-` `97``] ``+``=` `1` ` `  `    ``E ``=` `N ``/``/` `K  ` `    ``greaterE ``=` `[] ` `    ``lessE ``=` `[] ` ` `  `    ``for` `i ``in` `range``(``0``, ``26``):  ` ` `  `        ``# Two arrays with number of  ` `        ``# operations required  ` `        ``if` `count[i] < E: ` `            ``lessE.append(E ``-` `count[i])  ` `        ``else``: ` `            ``greaterE.append(count[i] ``-` `E)  ` ` `  `    ``greaterE.sort()  ` `    ``lessE.sort()  ` ` `  `    ``mi ``=` `float``(``'inf'``)  ` `    ``for` `i ``in` `range``(``0``, K ``+` `1``):  ` ` `  `        ``# Checking for all possibility  ` `        ``set1, set2 ``=` `i, K ``-` `i  ` ` `  `        ``if` `(``len``(greaterE) >``=` `set1 ``and`  `            ``len``(lessE) >``=` `set2):  ` ` `  `            ``step1, step2 ``=` `0``, ``0` ` `  `            ``for` `j ``in` `range``(``0``, set1):  ` `                ``step1 ``+``=` `greaterE[j]  ` ` `  `            ``for` `j ``in` `range``(``0``, set2):  ` `                ``step2 ``+``=` `lessE[j]  ` ` `  `            ``mi ``=` `min``(mi, ``max``(step1, step2)) ` ` `  `    ``print``(mi)  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``S ``=` `"accb"` `    ``N ``=` `len``(S)  ` `    ``K ``=` `2` ` `  `    ``minOperation(S, N, K)  ` `     `  `# This code is contributed by Rituraj Jain `

## C#

 `// C# program to convert the given string ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to find the minimum number of ` `    ``// operations to convert the given string ` `    ``static` `void` `minOperation(``string` `S, ``int` `N, ``int` `K) ` `    ``{ ` `        ``// Check if N is divisible by K ` `        ``if` `(N % K != 0)  ` `        ``{ ` `            ``Console.WriteLine(``"Not Possible"``); ` `        ``} ` `        ``else` `        ``{ ` `            ``// Array to store frequency of characters ` `            ``// in given string ` `            ``int` `[] count = ``new` `int``[26]; ` `             `  `            ``for` `(``int` `i = 0; i < N; i++)  ` `            ``{ ` `                ``count[(S[i] - 97)]++; ` `            ``} ` `         `  `            ``int` `E = N / K; ` `         `  `            ``List<``int``> greaterE = ``new` `List<``int``>(); ` `            ``List<``int``> lessE = ``new` `List<``int``>(); ` `             `  `            ``for` `(``int` `i = 0; i < 26; i++)  ` `            ``{ ` `         `  `                ``// Two arrays with number of operations ` `                ``// required ` `                ``if` `(count[i] < E) ` `                    ``lessE.Add(E - count[i]); ` `                ``else` `                    ``greaterE.Add(count[i] - E); ` `            ``} ` `         `  `            ``greaterE.Sort(); ` `            ``lessE.Sort(); ` `         `  `            ``int` `mi = Int32.MaxValue; ` `         `  `            ``for` `(``int` `i = 0; i <= K; i++)  ` `            ``{ ` `         `  `                ``// Checking for all possibility ` `                ``int` `set1 = i; ` `                ``int` `set2 = K - i; ` `         `  `                ``if` `(greaterE.Count >= set1 &&  ` `                            ``lessE.Count >= set2)  ` `                ``{ ` `         `  `                    ``int` `step1 = 0; ` `                    ``int` `step2 = 0; ` `         `  `                    ``for` `(``int` `j = 0; j < set1; j++) ` `                        ``step1 += greaterE[j]; ` `         `  `                    ``for` `(``int` `j = 0; j < set2; j++) ` `                        ``step2 += lessE[j]; ` `         `  `                    ``mi = Math.Min(mi, Math.Max(step1, step2)); ` `                ``} ` `            ``} ` `         `  `            ``Console.WriteLine(mi); ` `        ``} ` ` `  `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `        ``string` `S = ``"accb"``; ` `        ``int` `N = S.Length; ` `        ``int` `K = 2; ` `     `  `        ``minOperation(S, N, K); ` `    ``} ` `} ` ` `  `// This code is contributed by ihritik `

Output:

```1
```

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