0349MediumArrayTwo PointersBinary SearchSliding WindowSortingHeap (Priority Queue)

Nearest Gauge Marks

Tracked in this browser only

Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A calibration bench carries gauge marks etched along a rail. Their readings are listed in arr from lowest to highest, and the same reading may be etched more than once. A technician aims the probe at reading x and wants the k marks nearest to that aim.

Mark a is nearer than mark b when |a - x| is smaller than |b - x|, and when the two distances are equal the smaller reading is treated as the nearer one.

Return the k chosen readings listed from lowest to highest. When a reading is etched several times, each etching counts as its own mark.

Examples

Example 1

Input: arr = [2,5,9,13,20], k = 3, x = 10
Output: [5,9,13]

The distances from 10 are 8, 5, 1, 3 and 10. The three smallest belong to 9, 13 and 5, and the answer lists those readings from lowest to highest.

Example 2

Input: arr = [1,3,5,7], k = 1, x = 4
Output: [3]

Both 3 and 5 sit one away from 4. The tie goes to the smaller reading, so 3 is returned.

Example 3

Input: arr = [-9,-3,0,4], k = 3, x = -2
Output: [-3,0,4]

The distances are 7, 1, 2 and 6, so -3, 0 and 4 are the three nearest and -9 is left out.

Constraints

1 <= k <= arr.length
1 <= arr.length <= 10^4
arr is listed from lowest to highest.
-10^4 <= arr[i] <= 10^4
-10^4 <= x <= 10^4

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def find_closest_elements(arr: list[int], k: int, x: int) -> list[int]:

Java

public List<Integer> findClosestElements(int[] arr, int k, int x)

The intended approach

Time O(log(n - k) + k)Space O(k) for the returned run

The chosen marks always form one run of `k` neighbouring entries, so search for the run's first index between 0 and `len(arr) - k`. For a candidate start, compare the distance from `x` to the mark at that start against the distance to the mark just past the run's end: the farther of the two says which side to drop, and equal distances keep the lower start so the tie rule is honoured. Slice the run once the search settles.

Pythonentry point: find_closest_elements

def find_closest_elements(arr, k, x):
    # Every answer is a window of k neighbouring marks, so the only question is where
    # the window starts.  Comparing the mark just outside the left edge against the
    # mark just outside the right edge tells which half of the remaining starts to
    # discard, so the start can be binary searched.
    # Time O(log(n - k) + k).  Space O(k) for the returned window.
    lo = 0
    hi = len(arr) - k
    while lo < hi:
        mid = (lo + hi) // 2
        if x - arr[mid] > arr[mid + k] - x:
            lo = mid + 1
        else:
            hi = mid
    return arr[lo:lo + k]

The obvious version

Order every mark by distance from `x`, breaking equal distances by the smaller reading, keep the first `k` of that order, and sort those back into reading order.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_nearest_marks(arr, k, x):
    ranked = sorted(arr, key=lambda v: (abs(v - x), v))
    chosen = ranked[:k]
    chosen.sort()
    return chosen

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in