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Formatted question description: https://leetcode.ca/all/1060.html

1060. Missing Element in Sorted Array

Level

Medium

Description

Given a sorted array A of unique numbers, find the K-th missing number starting from the leftmost number of the array.

Example 1:

Input: A = [4,7,9,10], K = 1

Output: 5

Explanation:

The first missing number is 5.

Example 2:

Input: A = [4,7,9,10], K = 3

Output: 8

Explanation:

The missing numbers are [5,6,8,…], hence the third missing number is 8.

Example 3:

Input: A = [1,2,4], K = 3

Output: 6

Explanation:

The missing numbers are [3,5,6,7,…], hence the third missing number is 6.

Note:

  1. 1 <= A.length <= 50000
  2. 1 <= A[i] <= 1e7
  3. 1 <= K <= 1e8

Solution

If the array nums is empty, simply return k since all numbers are missing in nums.

If k is greater than or equal to the last number in nums, then only the numbers in nums are not missing and all the other numbers are missing, so the k-th missing number is k + nums[0] + nums.length - 1.

Otherwise, starting from the first letter in nums, check each number that is greater, and if a number is not in the array, then it is missing and decrease k by 1. The process ends until the end of nums is reached or k becomes 0. Finally, return the current number plus k.

  • class Solution {
    public int missingElement(int[] nums, int k) {
        if (nums == null || nums.length == 0)
            return k;
        int length = nums.length;
        if (k >= nums[length - 1])
            return k + nums[0] + length - 1;
        int num = nums[0];
        int index = 1;
        while (index < length && k > 0) {
            num++;
            if (num == nums[index])
                index++;
            else
                k--;
        }
        return num + k;
    }
}

  • class Solution {
public:
    int missingElement(vector<int>& nums, int k) {
        auto missing = [&](int i) {
            return nums[i] - nums[0] - i;
        };
        int n = nums.size();
        if (k > missing(n - 1)) {
            return nums[n - 1] + k - missing(n - 1);
        }
        int l = 0, r = n - 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (missing(mid) >= k) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return nums[l - 1] + k - missing(l - 1);
    }
};

  • class Solution:
    def missingElement(self, nums: List[int], k: int) -> int:
        def missing(i: int) -> int:
            return nums[i] - nums[0] - i

        n = len(nums)
        if k > missing(n - 1):
            return nums[n - 1] + k - missing(n - 1)
        l, r = 0, n - 1
        while l < r:
            mid = (l + r) >> 1
            if missing(mid) >= k:
                r = mid
            else:
                l = mid + 1
        return nums[l - 1] + k - missing(l - 1)


  • func missingElement(nums []int, k int) int {
	missing := func(i int) int {
		return nums[i] - nums[0] - i
	}
	n := len(nums)
	if k > missing(n-1) {
		return nums[n-1] + k - missing(n-1)
	}
	l, r := 0, n-1
	for l < r {
		mid := (l + r) >> 1
		if missing(mid) >= k {
			r = mid
		} else {
			l = mid + 1
		}
	}
	return nums[l-1] + k - missing(l-1)
}

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