Find the element that appears once

最新推荐文章于 2025-01-13 15:16:10 发布

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最新推荐文章于 2025-01-13 15:16:10 发布

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本文介绍了一种在时间复杂度为O(n)，空间复杂度为O(1)的情况下，解决给定数组中每个元素出现三次，仅有一个元素出现一次的问题的算法。通过逐位求和并取模的方法，可以找出那个只出现一次的元素。

reference:

http://www.geeksforgeeks.org/find-the-element-that-appears-once/

Problem Definition:

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

Solution:

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

Code:

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;
}