Find a special number in an array

Question

There are many numbers in an array and each number appears three times excepting for one special number appearing once. Here is the question: how can I find the special numb

Answer 1

Following is another O(n) time complexity and O(1) extra space method

suggested by aj. We can sum the bits in same positions for all the numbers and take modulo with 3.

The bits for which sum is not multiple of 3, are the bits of number with single occurrence. Let us consider

the example array {5, 5, 5, 8}.

The 101, 101, 101, 1000

Sum of first bits%3 = (1 + 1 + 1 + 0)%3 = 0;

Sum of second bits%3 = (0 + 0 + 0 + 0)%0 = 0;

Sum of third bits%3 = (1 + 1 + 1 + 0)%3 = 0;

Sum of fourth bits%3 = (1)%3 = 1;

Hence number which appears once is 1000

#include <stdio.h>
#define INT_SIZE 32

int getSingle(int arr[], int n)
{
// Initialize result
int result = 0;

int x, sum;

// Iterate through every bit
for (int i = 0; i < INT_SIZE; i++)
{
  // Find sum of set bits at ith position in all
  // array elements
  sum = 0;
  x = (1 << i);
  for (int j=0; j< n; j++ )
  {
      if (arr[j] & x)
        sum++;
  }

  // The bits with sum not multiple of 3, are the
  // bits of element with single occurrence.
  if (sum % 3)
    result |= x;
}

return result;
}

// Driver program to test above function
int main()
{
int arr[] = {12, 1, 12, 3, 12, 1, 1, 2, 3, 2, 2, 3, 7};
int n = sizeof(arr) / sizeof(arr[0]);
printf("The element with single occurrence is %d ",getSingle(arr, n));
return 0;
}

Answer 2

Since nobody's saying it, I will: hashtable.

You can calculate how many times each element occurs in the array in O(n) with simple hashtable (or hashmap).

Answer 3

A possible algorithm (very generic, not tested) :

function findMagicNumber(arr[0...n])
   magic_n := NaN

   if n = 1 then
      magic_n := arr[0]
   else if n > 1 then
      quicksort(arr)

      old_n := arr[0]
      repeat := 0

      for i := 1 to n
         cur_n := arr[i]
         repeat := repeat + 1
         if cur_n ≠ old_n then
            if repeat = 1 then
               magic_n := old_n
            old_n := cur_n
            repeat := 0

   return magic_n