2320 Find All K Distant Indices In An Array

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|  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

class Solution:
    def findKDistantIndices(self, nums: List[int], key: int, k: int) -> List[int]:
        arr =[]

        for i, m in enumerate(nums):
            if m == key:
                arr.append(i)
        res = set()
        for m in arr:
            for n in range(-k, k+1):
                if m + n >= 0 and m + n < len(nums):
                    res.add(m+n)
        return list(res)