2200. Find All K-Distant Indices in an Array

#### QUESTION:

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 {
func findKDistantIndices(_ nums: [Int], _ key: Int, _ k: Int) -> [Int] {
var keyIndex:[Int] = []
for index in nums.indices {
if (nums[index] == key) {
keyIndex.append(index)
}
}
var result:[Int] = []
for indexI in nums.indices {
for indexJ in keyIndex {
if (abs(indexI - indexJ) <= k) {
result.append(indexI)
break
}
}
}
return result.sorted()
}
}
``````
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