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268. Missing Number

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 <= 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?

Solutions

Solution 1: Bitwise Operation

The XOR operation has the following properties:

  • Any number XOR 0 is still the original number, i.e., $x \oplus 0 = x$;
  • Any number XOR itself is 0, i.e., $x \oplus x = 0$;

Therefore, we can traverse the array, perform XOR operation between each element and the numbers $[0,..n]$, and the final result will be the missing number.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

Solution 2: Mathematics

We can also solve this problem using mathematics. By calculating the sum of $[0,..n]$, subtracting the sum of all numbers in the array, we can obtain the missing number.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

  • class Solution {
    public int missingNumber(int[] nums) {
        int n = nums.length;
        int ans = n;
        for (int i = 0; i < n; ++i) {
            ans ^= (i ^ nums[i]);
        }
        return ans;
    }
}

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

  • class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        return reduce(xor, (i ^ v for i, v in enumerate(nums, 1)))


  • func missingNumber(nums []int) (ans int) {
	n := len(nums)
	ans = n
	for i, v := range nums {
		ans ^= (i ^ v)
	}
	return
}

  • function missingNumber(nums: number[]): number {
    const n = nums.length;
    let ans = n;
    for (let i = 0; i < n; ++i) {
        ans ^= i ^ nums[i];
    }
    return ans;
}


  • /**
 * @param {number[]} nums
 * @return {number}
 */
var missingNumber = function (nums) {
    const n = nums.length;
    let ans = n;
    for (let i = 0; i < n; ++i) {
        ans ^= i ^ nums[i];
    }
    return ans;
};


  • class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer
     */
    function missingNumber($nums) {
        $n = count($nums);
        $sumN = (($n + 1) * $n) / 2;
        for ($i = 0; $i < $n; $i++) {
            $sumN -= $nums[$i];
        }
        return $sumN;
    }
}

  • impl Solution {
    pub fn missing_number(nums: Vec<i32>) -> i32 {
        let n = nums.len() as i32;
        let mut ans = n;
        for (i, v) in nums.iter().enumerate() {
            ans ^= (i as i32) ^ v;
        }
        ans
    }
}

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