Leetcode - 137. Single Number II

Problem Description:
Given an array of integers, every element appears three times except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

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Analysis:
we could use bit manipulation to solve the problem.
00, 01, 10 represent 3 states, which for occurrence of 1, 2, 3 times of the incoming bit .

cur (a,b) incoming(c) next (a,b) 00 0 00 01 0 01 10 0 10 00 1 01 01 1 10 10 1 00

We could find the equation for a.
consider the case that a = 1.
we can conclude that
a = a &~b & ~c + ~a & b & c.
The same way :
b = ~a & b & ~c + ~a & ~b &c

we need to find the except one as the question don’t mention whether the one appears one or two time. so for ab
01 , 10 ==> 1 time
00 ==> 0 time
we should return a|b;

The above example is using bit. but when we actually do , we receive the number, for int, it’s 32 -bit, but for each bit we will do the same logic. For this operation, we can do it in parallel which is bit operation.

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The more details you can see this link

class Solution {public:    int singleNumber(vector<int>& nums) {    int a = 0, b = 0;    for (auto c : nums)    {        int tmp = (a & ~b & ~c) + (~a & b &c);        b = (~a & b & ~c) + (~a & ~b &c);        a = tmp;    }    return a | b;    }};

Here is a regular solution:

int singleNumber(int arr[], int n) {        // Initialize result    int result = 0;    int x, sum;    // Iterate through every bit    for (int i = 0; i < 32; 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;    }