# 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.

 // 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 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 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# 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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Improved By : vt_m, chitranayal

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