1146EasyArrayTwo Pointers

Positions Near a Given Reading

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Trains the technique from

LeetCode 2200Find All K-Distant Indices in an Array

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 row of readings is given, together with a value key that turns up somewhere in the row and a whole number k.

A position i is near when the row holds key at some position j with the gap between i and j no more than k.

Return every near position, in increasing order.

Examples

Example 1

Input: readings = [5, 1, 1, 1, 5], key = 5, k = 1
Output: [0, 1, 3, 4]

The value sits at each end of the row, so both ends and their single neighbours are near, while the middle position is two steps from either end.

Example 2

Input: readings = [4, 1, 1, 1, 1, 1, 1, 4], key = 4, k = 2
Output: [0, 1, 2, 5, 6, 7]

The value sits at each end again, and reaching two steps inwards from each covers the first three and the last three positions.

Example 3

Input: readings = [7, 7, 7], key = 7, k = 3
Output: [0, 1, 2]

Every position holds the value, so every position is near.

Constraints

1 <= readings.length <= 1000
1 <= readings[i] <= 1000
key turns up somewhere in readings
1 <= k <= readings.length

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_k_distant_indices(readings: list[int], key: int, k: int) -> list[int]:

Java

public List<Integer> findKDistantIndices(int[] readings, int key, int k)

The intended approach

Time O(n * k) for the marking plus O(n) to read it back.Space O(n) for the flags.

Each place the row holds `key` makes a whole band of positions near, reaching `k` either side of it and cut off at the ends of the row. So the answer is the union of those bands. Building it by asking, for each position, whether some `key` lies within reach costs the row length squared. Instead sweep the row once, and whenever `key` turns up mark its whole band as near. Bands from different places overlap freely, and marking simply writes the same flag again, so no special handling is needed. Reading the flags back in position order then gives the answer already sorted, with no position repeated however many bands covered it.

Pythonentry point: find_k_distant_indices

def find_k_distant_indices(readings, key, k):
    n = len(readings)
    near = [False] * n
    for j in range(n):
        if readings[j] == key:
            lo = j - k
            if lo < 0:
                lo = 0
            hi = j + k
            if hi > n - 1:
                hi = n - 1
            for i in range(lo, hi + 1):
                near[i] = True
    out = []
    for i in range(n):
        if near[i]:
            out.append(i)
    return out

The obvious version

Take each position in turn and sweep the whole row looking for a place holding `key` within reach, keeping the position the moment one turns up. Costs the row length squared.

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_find_k_distant_indices(readings, key, k):
    n = len(readings)
    out = []
    for i in range(n):
        for j in range(n):
            if readings[j] == key:
                gap = i - j if i > j else j - i
                if gap <= k:
                    out.append(i)
                    break
    return out

How this problem was checked

Reference solution executed against 15 test cases — 15 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

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