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Missing Number

Problem​

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.

Solution​

/**
 * @param {number[]} nums
 * @return {number}
 */
var missingNumber = function(nums) {
    const map = Array(nums.length + 1).fill(false);
    for (const n of nums) {
        map[n] = true;
    }
    return map.indexOf(false);
};

Use a hashmap map to keep track of which values in the range [0, n] appeared before with true, and return the index of the only entry that is false.

Follow-up​

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

Solution​

/**
 * @param {number[]} nums
 * @return {number}
 */
var missingNumber = function(nums) {
    const sum1 = (nums.length * (nums.length + 1)) / 2;
    const sum2 = nums.reduce((prev, acc) => prev + acc, 0);
    return sum1 - sum2;
};

The missing value is the sum of values in [0, n] minus the sum of values in nums.