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 <= A.length <= 500001 <= A[i] <= 1e71 <= 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;
}
}