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Missing Number⚓︎

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Description⚓︎

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.

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.

Constraints:

  • n == nums.length
  • 1 <= n <= 10^4
  • 0 <= nums[i] <= n
  • All the numbers of nums are unique.

Solution⚓︎

Way 1⚓︎

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class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int n = nums.size();
        int current = 0, sum = n * (n + 1) / 2;
        for (int x : nums) current += x;
        return sum - current;
    }
};
  • Time complexity: \(O(n)\);
  • Space complexity: \(O(1)\).

Way 2: XOR⚓︎

The overall XOR of an array is x, and the XOR of a part of the whole is y, then the XOR of the remaining part is x ^ y.

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class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int xorAll = 0, xorNums = 0;
        for (int i = 0; i < nums.size(); i++) {
            xorAll ^= i;
            xorNums ^= nums[i];
        }
        xorAll ^= nums.size();

        return xorAll ^ xorNums;
    }
};
  • Time complexity: \(O(n)\);
  • Space complexity: \(O(1)\).