# Minimum replacements to make elements of a ternary array same

Given a ternary array (every element has one the three possible values 1, 2 and 3). Our task is to replace the minimum number of numbers in it so that all the numbers in the array are equal to each other.

Examples:

```Input :  arr[] = 1 3 2 2 2 1 1 2 3
Output : 5
In this example, frequency of 1 is 3,
frequency of 2 is 4 and frequency of 3
is 2. As we can see that 2 is having the
more frequency than 1 and 3. So, if we
replace all the 1's and 3's by 2 then,
the resultant array has all the elements
equal to each other in minimum replacements.
Here, total no. of 1's and 3's is 5 so it
takes 5 replacements to replace them by 2.
Hence, the output is 5.

Input : arr[] = 3 3 2 2 1 3
Output : 3
In this example, 3 has the max frequency.
Hence, minimum number of replacements are
3 to replace 1 and 2 by 3. Hence, the output
is 3.
```

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

The approach is to calculate frequency of each element of the given array. Then, the difference of n(no. of elements) and max_frequency(frequency of the element occurs maximum time in the array) will be minimum number of replacements needed.

## C++

 `// CPP program minimum number of replacements ` `// needed to be performed to make all the numbers ` `// in the given array equal. ` `#include ` `using` `namespace` `std; ` ` `  `int` `minReplacements(``int` `arr[], ``int` `n) ` `{ ` `    ``// Find the most frequent element ` `    ``int` `freq[3] = { 0 }; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``freq[arr[i] - 1]++; ` `    ``int` `max_freq = *max_element(freq, freq + 3); ` ` `  `    ``// Returning count of replacing other elements ` `    ``// with the most frequent. ` `    ``return` `(n - max_freq); ` `} ` ` `  `// Driver Function ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 3, 2, 2, 2, 1, 1, 2, 3 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``cout << minReplacements(arr, n) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program minimum  ` `// number of replacements  ` `// needed to be performed  ` `// to make all the numbers  ` `// in the given array equal ` `import` `java .io.*; ` `import` `java .util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function for  ` `// minimum replacements ` `static` `int` `minReplacements(``int` `[]arr,  ` `                           ``int` `n) ` `{ ` `    ``// Find the most ` `    ``// frequent element ` `    ``int` `[]freq = ``new` `int``[``3``]; ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``freq[arr[i] - ``1``]++; ` `        ``Arrays.sort(freq); ` `    ``int` `max_freq = freq[``2``]; ` ` `  `    ``// Returning count of  ` `    ``// replacing other elements  ` `    ``// with the most frequent ` `    ``return` `(n - max_freq); ` `} ` ` `  `// Driver code ` `static` `public` `void` `main (String[] args) ` `{ ` `    ``int` `[]arr = {``1``, ``3``, ``2``, ``2``,  ` `                 ``2``, ``1``, ``1``, ``2``, ``3``}; ` `    ``int` `n = arr.length; ` `    ``System.out.println(minReplacements(arr, n)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by anuj_67. `

## Python 3

 `# Python 3 program minimum number of  ` `# replacements needed to be performed  ` `# to make all the numbers in the given ` `# array equal. ` ` `  `def` `minReplacements(arr, n): ` ` `  `    ``# Find the most frequent element ` `    ``freq ``=` `[``0``] ``*` `3` `    ``for` `i ``in` `range``(n): ` `        ``freq[arr[i] ``-` `1``] ``+``=` `1` `    ``freq.sort() ` `    ``max_freq ``=` `freq[``2``] ` ` `  `    ``# Returning count of replacing other  ` `    ``# elements with the most frequent. ` `    ``return` `(n ``-` `max_freq) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``arr ``=` `[ ``1``, ``3``, ``2``, ``2``,  ` `            ``2``, ``1``, ``1``, ``2``, ``3` `] ` `    ``n ``=` `len``(arr) ` `    ``print``( minReplacements(arr, n) ) ` ` `  `# This code is contributed  ` `# by ChitraNayal `

## C#

 `// C# program minimum number of ` `// replacements needed to be  ` `// performed to make all the ` `// numbers in the given array equal ` `using` `System; ` `using` `System.Linq; ` ` `  `public` `class` `GFG  ` `{ ` ` `  `// Function for minimum replacements ` `static` `int` `minReplacements(``int` `[]arr, ``int` `n) ` `{ ` `    ``// Find the most frequent element ` `    ``int` `[]freq = ``new` `int``[3]; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``freq[arr[i] - 1]++; ` `    ``int` `max_freq = freq.Max(); ` ` `  `    ``// Returning count of replacing other ` `    ``// elements with the most frequent ` `    ``return` `(n - max_freq); ` `} ` ` `  `    ``// Driver code ` `    ``static` `public` `void` `Main () ` `    ``{ ` `        ``int` `[]arr = {1, 3, 2, 2, 2, 1, 1, 2, 3}; ` `        ``int` `n = arr.Length; ` `        ``Console.WriteLine(minReplacements(arr, n)); ` `     `  `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

Output:

```5
```

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