Find the element that appears once

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martin_liang

最新推荐文章于 2025-01-13 15:16:10 发布

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本文介绍了一种在数组中查找仅出现一次的元素的方法，该元素在其他所有元素都出现三次的情况下。此方法的时间复杂度为O(n)，额外空间复杂度为O(1)。通过遍历每个位并对相同位置的比特求和后取模3，可以找到目标元素。

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Given an array where every element occurs three times, except one element which occurs only once. Find the element that occurs once. Expected time complexity is O(n) and O(1) extra space.
Examples:

Input: arr[] = {12, 1, 12, 3, 12, 1, 1, 2, 3, 3}
Output: 2

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;
}