# Minimum number of additons to make the string balanced

Given a string str of lowercase characters, the task is to find the minimum number of characters that need to added to the string in order to make it balanced. A string is said to be balanced if and only if the number of occurrences of each of the characters is equal.

Examples:

Input: str = “geeksforgeeks”
Output: 15
Add 2 ‘g’, 2 ‘k’, 2 ‘s’, 3 ‘f’, 3 ‘o’ and 3 ‘r’.

Input: str = “abcd”
Output: 0

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

Approach: In order to minimize the additions required, every character’s frequency must be made equal to the frequency of the element occurring most frequently. So first, create a frequency array and find the frequency of all the characters of the given string. Now, the required answer will be the sum of the absolute differences of the frequency of every character with the maximum frequency from the frequency array.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `#define MAX 26 ` ` `  `// Function to return the minimum additions ` `// required to balance the given string ` `int` `minimumAddition(string str, ``int` `len) ` `{ ` ` `  `    ``// To store the frequency of ` `    ``// the characters of str ` `    ``int` `freq[MAX] = { 0 }; ` ` `  `    ``// Update the frequency of the characters ` `    ``for` `(``int` `i = 0; i < len; i++) { ` `        ``freq[str[i] - ``'a'``]++; ` `    ``} ` ` `  `    ``// To store the maximum frequency from the array ` `    ``int` `maxFreq = *max_element(freq, freq + MAX); ` ` `  `    ``// To store the minimum additions required ` `    ``int` `minAddition = 0; ` `    ``for` `(``int` `i = 0; i < MAX; i++) { ` ` `  `        ``// Every character's frequency must be ` `        ``// equal to the frequency of the most ` `        ``// frequently occurring character ` `        ``if` `(freq[i] > 0) { ` `            ``minAddition += ``abs``(maxFreq - freq[i]); ` `        ``} ` `    ``} ` ` `  `    ``return` `minAddition; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"geeksforgeeks"``; ` `    ``int` `len = str.length(); ` ` `  `    ``cout << minimumAddition(str, len); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG  ` `{ ` `    ``final` `static` `int` `MAX = ``26``;  ` `     `  `    ``static` `int` `max_element(``int` `freq[]) ` `    ``{ ` `        ``int` `max_ele = freq[``0``]; ` `        ``for``(``int` `i = ``0``; i < MAX; i++) ` `        ``{ ` `            ``if``(max_ele < freq[i]) ` `                ``max_ele = freq[i]; ` `        ``} ` `        ``return` `max_ele; ` `    ``} ` `     `  `    ``// Function to return the minimum additions  ` `    ``// required to balance the given string  ` `    ``static` `int` `minimumAddition(String str, ``int` `len)  ` `    ``{  ` `     `  `        ``// To store the frequency of  ` `        ``// the characters of str  ` `        ``int` `freq[] = ``new` `int``[MAX]; ` `     `  `        ``// Update the frequency of the characters  ` `        ``for` `(``int` `i = ``0``; i < len; i++)  ` `        ``{  ` `            ``freq[str.charAt(i) - ``'a'``]++;  ` `        ``}  ` `     `  `        ``// To store the maximum frequency from the array  ` `        ``int` `maxFreq = max_element(freq);  ` `     `  `        ``// To store the minimum additions required  ` `        ``int` `minAddition = ``0``;  ` `        ``for` `(``int` `i = ``0``; i < MAX; i++)  ` `        ``{  ` `     `  `            ``// Every character's frequency must be  ` `            ``// equal to the frequency of the most  ` `            ``// frequently occurring character  ` `            ``if` `(freq[i] > ``0``)  ` `            ``{  ` `                ``minAddition += Math.abs(maxFreq - freq[i]);  ` `            ``}  ` `        ``}  ` `        ``return` `minAddition;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``String str = ``"geeksforgeeks"``;  ` `        ``int` `len = str.length();  ` `     `  `        ``System.out.println(minimumAddition(str, len));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach ` `MAX` `=` `26` ` `  `# Function to return the minimum additions ` `# required to balance the given str1ing ` `def` `minimumAddition(str1, ``Len``): ` ` `  `    ``# To store the frequency of ` `    ``# the characters of str1 ` `    ``freq ``=` `[``0` `for` `i ``in` `range``(``MAX``)] ` ` `  `    ``# Update the frequency of the characters ` `    ``for` `i ``in` `range``(``Len``): ` `        ``freq[``ord``(str1[i]) ``-` `ord``(``'a'``)] ``+``=` `1` ` `  `    ``# To store the maximum frequency from the array ` `    ``maxFreq ``=` `max``(freq) ` ` `  `    ``# To store the minimum additions required ` `    ``minAddition ``=` `0` `    ``for` `i ``in` `range``(``MAX``): ` ` `  `        ``# Every character's frequency must be ` `        ``# equal to the frequency of the most ` `        ``# frequently occurring character ` `        ``if` `(freq[i] > ``0``): ` `            ``minAddition ``+``=` `abs``(maxFreq ``-` `freq[i]) ` ` `  `    ``return` `minAddition ` ` `  `# Driver code ` `str1 ``=` `"geeksforgeeks"` `Len` `=` `len``(str1) ` ` `  `print``(minimumAddition(str1, ``Len``)) ` ` `  `# This code is contributed Mohit Kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GFG  ` `{ ` `    ``static` `int` `MAX = 26;  ` `     `  `    ``static` `int` `max_element(``int` `[]freq) ` `    ``{ ` `        ``int` `max_ele = freq; ` `        ``for``(``int` `i = 0; i < MAX; i++) ` `        ``{ ` `            ``if``(max_ele < freq[i]) ` `                ``max_ele = freq[i]; ` `        ``} ` `        ``return` `max_ele; ` `    ``} ` `     `  `    ``// Function to return the minimum additions  ` `    ``// required to balance the given string  ` `    ``static` `int` `minimumAddition(String str, ``int` `len)  ` `    ``{  ` `     `  `        ``// To store the frequency of  ` `        ``// the characters of str  ` `        ``int` `[]freq = ``new` `int``[MAX]; ` `     `  `        ``// Update the frequency of the characters  ` `        ``for` `(``int` `i = 0; i < len; i++)  ` `        ``{  ` `            ``freq[str[i] - ``'a'``]++;  ` `        ``}  ` `     `  `        ``// To store the maximum frequency from the array  ` `        ``int` `maxFreq = max_element(freq);  ` `     `  `        ``// To store the minimum additions required  ` `        ``int` `minAddition = 0;  ` `        ``for` `(``int` `i = 0; i < MAX; i++)  ` `        ``{  ` `     `  `            ``// Every character's frequency must be  ` `            ``// equal to the frequency of the most  ` `            ``// frequently occurring character  ` `            ``if` `(freq[i] > 0)  ` `            ``{  ` `                ``minAddition += Math.Abs(maxFreq - freq[i]);  ` `            ``}  ` `        ``}  ` `        ``return` `minAddition;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main (String[] args) ` `    ``{  ` `        ``String str = ``"geeksforgeeks"``;  ` `        ``int` `len = str.Length;  ` `     `  `        ``Console.WriteLine(minimumAddition(str, len));  ` `    ``}  ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```15
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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