Formatted question description: https://leetcode.ca/all/2200.html

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

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

Similar Questions:

## Solution 1. Two Pointers

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

// Time: O(N)
// Space: O(N)
class Solution {
public:
vector<int> findKDistantIndices(vector<int>& A, int key, int k) {
int N = A.size();
vector<int> ans, idx;
for (int i = 0; i < N; ++i){
if (A[i] == key) idx.push_back(i); // idx is a list of indices whose corresponding value is key.
}
for (int i = 0, j = 0; i < N && j < idx.size(); ++i) {
if (i < idx[j] - k) continue; // If i is not yet in range of the next key element at idx[j], skip.
while (j < idx.size() && i > idx[j] + k) ++j; // While i > idx[j] + k, keep incrementing j to bring idx[j] in range of i.
if (j < idx.size() && i <= idx[j] + k && i >= idx[j] - k) ans.push_back(i); // add i to the answer if idx[j] - k <= i <= idx[j] + k.
}
return ans;
}
};

• // Todo

• # 2200. Find All K-Distant Indices in an Array
# https://leetcode.com/problems/find-all-k-distant-indices-in-an-array/

class Solution:
def findKDistantIndices(self, nums: List[int], key: int, k: int) -> List[int]:
n = len(nums)
indices = [i for i, x in enumerate(nums) if x == key]
res = []

for i in range(n):
for j in indices:
if abs(i - j) <= k:
res.append(i)
break

return res



## Solution 2. Two Pointers

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

// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> findKDistantIndices(vector<int>& A, int key, int k) {
int N = A.size(), j = 0;
vector<int> ans;
for (int i = 0, j = 0; i < N; ++i) {
while (j < N && (A[j] != key || j < i - k)) ++j; // Find the first index j that A[j] == key and j >= i - k.
if (j == N) break;
if (i <= j + k && i >= j - k) ans.push_back(i); // add i to answer if j - k <= i <= j + k.
}
return ans;
}
};

• // Todo

• # 2200. Find All K-Distant Indices in an Array
# https://leetcode.com/problems/find-all-k-distant-indices-in-an-array/

class Solution:
def findKDistantIndices(self, nums: List[int], key: int, k: int) -> List[int]:
n = len(nums)
indices = [i for i, x in enumerate(nums) if x == key]
res = []

for i in range(n):
for j in indices:
if abs(i - j) <= k:
res.append(i)
break

return res



## Discuss

https://leetcode.com/problems/find-all-k-distant-indices-in-an-array/discuss/1845499