Missing Number LeetCode Solution

Problem – Missing Number

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 <= 104
  • 0 <= nums[i] <= n
  • All the numbers of nums are unique.

Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?

Missing Number LeetCode Solution in Java

public int missingNumber(int[] nums) {

    int xor = 0, i = 0;
	for (i = 0; i < nums.length; i++) {
		xor = xor ^ i ^ nums[i];
	}

	return xor ^ i;
}

Missing Number LeetCode Solution in C++

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int result = nums.size();
        int i=0;
        
        for(int num:nums){
            result ^= num;
            result ^= i;
            i++;
        }
        
        return result;
    }
};

Missing Number LeetCode Solution in Python

class Solution(object):
    def missingNumber(self, nums):
        return sum(range(len(nums)+1)) - sum(nums)
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