---
description: >-
  Author: @wingkwong |
  https://leetcode.com/problems/find-all-k-distant-indices-in-an-array/
---

# 2200 - Find All K-Distant Indices in an Array (Easy)

## Problem Link

https://leetcode.com/problems/find-all-k-distant-indices-in-an-array/

## Problem Statement

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`

## Approach 1: Brute Force

Just do what it says.

<SolutionAuthor name="@wingkwong"/>

```cpp
class Solution {
public:
    vector<int> findKDistantIndices(vector<int>& nums, int key, int k) {
        vector<int> ans;
        int n = nums.size();
        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;
    }
};
```
