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Abhishek Chaudhary

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Find All K-Distant Indices in an Array

You are given a 0-indexed integer array `nums` and two integers `key` and `k`. A k-distant index is an index `i` of `nums` for which there exists at least one index `j` such that `|i - j| <= k` and `nums[j] == key`.

Return a list of all k-distant indices sorted in increasing order.

Example 1:

Input: nums = [3,4,9,1,3,9,5], key = 9, k = 1
Output: [1,2,3,4,5,6]
Explanation: Here, `nums[2] == key` and `nums[5] == key. - For index 0, |0 - 2| > k and |0 - 5| > k, so there is no j` where `|0 - j| <= k` and `nums[j] == key. Thus, 0 is not a k-distant index. - For index 1, |1 - 2| <= k and nums[2] == key, so 1 is a k-distant index. - For index 2, |2 - 2| <= k and nums[2] == key, so 2 is a k-distant index. - For index 3, |3 - 2| <= k and nums[2] == key, so 3 is a k-distant index. - For index 4, |4 - 5| <= k and nums[5] == key, so 4 is a k-distant index. - For index 5, |5 - 5| <= k and nums[5] == key, so 5 is a k-distant index. - For index 6, |6 - 5| <= k and nums[5] == key, so 6 is a k-distant index.`
Thus, we return [1,2,3,4,5,6] which is sorted in increasing order.

Example 2:

Input: nums = [2,2,2,2,2], key = 2, k = 2
Output: [0,1,2,3,4]
Explanation: For all indices i in nums, there exists some index j such that |i - j| <= k and nums[j] == key, so every index is a k-distant index.
Hence, we return [0,1,2,3,4].

Constraints:

• `1 <= nums.length <= 1000`
• `1 <= nums[i] <= 1000`
• `key` is an integer from the array `nums`.
• `1 <= k <= nums.length`

SOLUTION:

``````class Solution:
def findKDistantIndices(self, nums: List[int], key: int, k: int) -> List[int]:
n = len(nums)
indices = set()
for i, el in enumerate(nums):
if el == key:
for j in range(max(i - k, 0), min(i + k + 1, n)):