703480. K-th missing element
Here is the Markdown blog post for the GeeksforGeeks problem "K-th missing element":
K-th Missing Element
Slug: k-th-missing-element3635 Difficulty: Easy Id: 703480 Topic Tags: Arrays, Data Structures Company Tags: Amazon, Facebook, Apple, Google
Summary
The K-th missing element problem involves finding the K-th missing element in a sequence of integers. The input consists of a sorted array of integers with some elements missing. The task is to find the K-th missing element. This problem requires an understanding of arrays and data structures.
Detailed Explanation
To solve this problem, we first need to understand what the problem is asking for. It's asking us to find the K-th missing element in a sorted array of integers with some elements missing. The key concept here is that the input array is sorted, which allows us to use a simple iteration-based approach.
Here is a step-by-step breakdown of the solution:
- Initialize two pointers,
leftandright, to the start and end of the array respectively. - Iterate through the array until
right - left + 1(the difference between the right and left pointers plus one) is equal to K+1. This means we have found the K-th missing element. - In each iteration, check if there are any elements missing between
leftandright. If there are, increment theleftpointer until we find a non-missing element. - Once we have found the K-th missing element, return it.
Here is an ASCII art diagram to illustrate this approach:
+---------------+
| 1, 2, 3, ... |
+---------------+
|
v
+---------------+
| ? (K-1st missing) |
+---------------+
|
v
+---------------+
| K-th missing |
+---------------+
Time complexity: O(n), where n is the length of the array. Space complexity: O(1).
Optimized Solutions
Java
public int findKthMissingElement(int[] arr, int k) {
int left = 0;
int right = arr.length - 1;
for (int i = 0; i < k; i++) {
while (left <= right && arr[left] + 1 != arr[right]) {
left++;
}
if (left > right) {
return arr[left - 1] + k - i;
}
right--;
}
return arr[right] + k;
}