0268. Missing Number

https://leetcode.com/problems/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.

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

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:

• n == nums.length

• 1 <= n <= 104

• 0 <= nums[i] <= n

• All the numbers of nums are unique.

ac1: make ust of index

class Solution {
    public int missingNumber(int[] nums) {
        // edge cases
        if (nums == null || nums.length == 0) throw new IllegalArgumentException("invalid input");

        // iterate
        for (int i = 0; i < nums.length; i++) {
            while (nums[i] < nums.length && nums[i] != i) {
                // swap
                int tmp = nums[i];
                nums[i] = nums[tmp];
                nums[tmp] = tmp;
            }
        }

        // check
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != i) return i;
        }

        return nums.length;
    }
}

ac2: math

class Solution {
    public int missingNumber(int[] nums) {
        // edge cases
        if (nums == null || nums.length == 0) throw new IllegalArgumentException("invalid input");

        // math
        int n = nums.length;
        int sumExpected = n * (n + 1) / 2;
        int sum = 0;
        for (int i : nums) {
            sum += i;
        }

        return sumExpected - sum;
    }
}

ac3: XOR

class Solution {
    public int missingNumber(int[] nums) {
        // edge cases
        if (nums == null || nums.length == 0) throw new IllegalArgumentException("invalid input");

        int res = 0 ^ nums[0];
        for (int i = 1; i < nums.length; i++) {
            res = res ^ i ^ nums[i];
        }

        return res ^ nums.length;
    }
}