658. Find k closest elements | Leetcode | Medium | Java | Binary Search | Two Pointers

When dealing with a sorted array and the objective is to identify elements closest to a given target, an optimized solution involves leveraging binary search to efficiently locate the first closest element. Following this, two pointers can be utilized to traverse the array and identify the remaining closest elements.

Binary Search for Closest Element:

Employ binary search to find the index of the element in the sorted array that is closest to the target.
If an exact match is not found, the binary search will pinpoint the immediate smaller element.

Two Pointers for Exploration:

Initialize two pointers, one pointing to the left and the other to the right of the identified closest element index.
Traverse both sides, comparing the absolute differences between the target and the elements at the pointers.
Select the element with the smaller absolute difference and move the corresponding pointer accordingly.
Continue this process until the desired number of closest elements is achieved.

It was asked in Amazon and Yandex in the last six months.

Other problems for practice:

[iamnoob]

My approach is also quite the same with Log(n) + k + k => BS + Find boundary using i and j pointer + add elements from i to j-1/j which is again O(k).

[satvikpandey]

the most optimal solution would be to use binary search + sliding window.

[udaykulkarni]

Great explanation : ) Radhe Radhe. Earned a sub! <3

[princekirar]

For test case :
[1, 2, 8] x = 7 && k = 1 will it work correctly ?