[Simple Algorithm] 49. Missing Numbers

topic:

Given a sequence of n numbers from 0 , 1 , 2 , ..., n , find the number of 0 .. n that does not appear in the sequence.

Case 1

Input: [ 3 , 0 , 1 ]
Output: 2
 

Case 2

Input: [ 9 , 6 , 4 , 2 , 3 , 5 , 7 , 0 , 1 ]
Output: 8
 

Precautions:
Your algorithm should run with linear complexity. Can you implement it using only constant extra space complexity?

Problem solving ideas:

XOR XOR from 0 to n to get the result val, and then XOR val with each element in the array to get the result. The solution to the subproblem is similar to a number from 1 to n where a number appears twice, and the solution to find the number is the same.

code show as below:

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int n = nums.size()+1;
        int res = 0;
        for(int i = 0;i < n; ++i){
            res ^=i;    
        }
        
        for(int i = 0;i < nums.size();++i){
            res ^=nums[i];
        }
        
        return res;
    }
};

Solution two: