268. 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.

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

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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.

Solution

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class Solution {
    public int missingNumber(int[] nums) {
        int n = nums.length;
        int sum = 0;
        for (int num: nums) {
            sum += num;
        }
        
        return ((n + 1) * n) / 2 - sum;
    }
    // 0 1 2 3 4 5 6 7 8 9
    // 10 * 9 / 2 = 45
}

Solution 2022-01-30

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class Solution {
    public int missingNumber(int[] nums) {
        int sum = 0;
        for (int num: nums) {
            sum += num;
        }
        int n = nums.length;
        return n * (n + 1) / 2 - sum;
    }
}