1060. Missing Element in Sorted Array

Description

Given an integer array nums which is sorted in ascending order and all of its elements are unique and given also an integer k, return the kth missing number starting from the leftmost number of the array.

 

Example 1:

Input: nums = [4,7,9,10], k = 1
Output: 5
Explanation: The first missing number is 5.

Example 2:

Input: nums = [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: nums = [1,2,4], k = 3
Output: 6
Explanation: The missing numbers are [3,5,6,7,...], hence the third missing number is 6.

 

Constraints:

  • 1 <= nums.length <= 5 * 104
  • 1 <= nums[i] <= 107
  • nums is sorted in ascending order, and all the elements are unique.
  • 1 <= k <= 108

 

Follow up: Can you find a logarithmic time complexity (i.e., O(log(n))) solution?

Solutions

Solution 1

Python Code
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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)

Java Code
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class Solution {
    public int missingElement(int[] nums, int k) {
        int n = nums.length;
        if (k > missing(nums, n - 1)) {
            return nums[n - 1] + k - missing(nums, n - 1);
        }
        int l = 0, r = n - 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (missing(nums, mid) >= k) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return nums[l - 1] + k - missing(nums, l - 1);
    }

    private int missing(int[] nums, int i) {
        return nums[i] - nums[0] - i;
    }
}

C++ Code
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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);
    }
};

Go Code
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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)
}