Missing Number

Tags : array, leetcode, cpp, easy

Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

Examples #

Example 1:

Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.

Example 2:

Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.

Example 3:

Input: nums = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.

Example 4:

Input: nums = [0]
Output: 1
Explanation: n = 1 since there is 1 number, so all numbers are in the range [0,1]. 1 is the missing number in the range since it does not appear in nums.

Constraints #

Solutions #

class Solution {
public:
    int missingNumber(vector<int>& nums) {
      // sort(nums.begin(), nums.end());
      // for(int i=0; i<nums.size(); i++) if(i!=nums[i]) return i;
      int res = 0;
      for(int i=0; i<nums.size(); i++){
        res ^= (i+1);
        res ^= nums[i];
      }
      return res;
    }
};