# 2200. Find All K-Distant Indices in an Array

## Description

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

## Solutions

Solution 1: Enumeration

 We enumerate the index $i$ in the range $[0, n)$, and for each index $i$, we enumerate the index $j$ in the range $[0, n)$. If $i - j \leq k$ and $nums[j] = key$, then $i$ is a K-nearest neighbor index. We add $i$ to the answer array, then break the inner loop and enumerate the next index $i$.

The time complexity is $O(n^2)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

Solution 2: Preprocessing + Binary Search

We can preprocess to get the indices of all elements equal to $key$, recorded in the array $idx$. All index elements in the array $idx$ are sorted in ascending order.

Next, we enumerate the index $i$. For each index $i$, we can use binary search to find elements in the range $[i - k, i + k]$ in the array $idx$. If there are elements, then $i$ is a K-nearest neighbor index. We add $i$ to the answer array.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

Solution 3: Two Pointers

We enumerate the index $i$, and use a pointer $j$ to point to the smallest index that satisfies $j \geq i - k$ and $nums[j] = key$. If $j$ exists and $j \leq i + k$, then $i$ is a K-nearest neighbor index. We add $i$ to the answer array.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

• class Solution {
public List<Integer> findKDistantIndices(int[] nums, int key, int k) {
int n = nums.length;
List<Integer> ans = new ArrayList<>();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (Math.abs(i - j) <= k && nums[j] == key) {
break;
}
}
}
return ans;
}
}

• class Solution {
public:
vector<int> findKDistantIndices(vector<int>& nums, int key, int k) {
int n = nums.size();
vector<int> ans;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (abs(i - j) <= k && nums[j] == key) {
ans.push_back(i);
break;
}
}
}
return ans;
}
};

• class Solution:
def findKDistantIndices(self, nums: List[int], key: int, k: int) -> List[int]:
ans = []
n = len(nums)
for i in range(n):
if any(abs(i - j) <= k and nums[j] == key for j in range(n)):
ans.append(i)
return ans


• func findKDistantIndices(nums []int, key int, k int) (ans []int) {
for i := range nums {
for j, x := range nums {
if abs(i-j) <= k && x == key {
ans = append(ans, i)
break
}
}
}
return ans
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• function findKDistantIndices(nums: number[], key: number, k: number): number[] {
const n = nums.length;
const ans: number[] = [];
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
if (Math.abs(i - j) <= k && nums[j] === key) {
ans.push(i);
break;
}
}
}
return ans;
}