Find Element with Different Frequency in C++

Suppose we have an array of N numbers, where each element in the array appears same number of times (m times, this is also given) except one element, We have to find this element.

So, if the input is like A = [6, 2, 7, 2, 2, 6, 6], m = 3, then the output will be 7.

To solve this, we will follow these steps −

• INT_SIZE := 8 * size of an integer type variable

• Define an array count of size − INT_SIZE. and fill with 0

• for initialize i := 0, when i < INT_SIZE, update (increase i by 1), do:

  • for initialize j := 0, when j < size, update (increase j by 1), do −

    • if (arr[j] AND 2^i) is not equal to 0, then −

      • count[i] := count[i] + 1

    • res := 0

  • for initialize i := 0, when i < INT_SIZE, update (increase i by 1), do −

    • res := res + ((count[i] mod m) * 2^i)

  • return res

Example 

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
int selectUnique(unsigned int arr[], int size, int m){
   int INT_SIZE = 8 * sizeof(unsigned int);
   int count[INT_SIZE];
   memset(count, 0, sizeof(count));
   for(int i = 0; i < INT_SIZE; i++)
      for(int j = 0; j < size; j++)
         if((arr[j] & (1 << i)) != 0)
            count[i] += 1;
   unsigned res = 0;
   for(int i = 0; i < INT_SIZE; i++)
      res += (count[i] % m) * (1 << i);
   return res;
}
main(){
   unsigned int arr[] = { 6, 2, 5, 2, 2, 6, 6 };
   int size = sizeof(arr) / sizeof(arr[0]);
   int m = 3;
   cout << selectUnique(arr, size, m);
}

Input

{ 6, 2, 5, 2, 2, 6, 6 }

Output

5