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 <= 10001 <= nums[i] <= 1000keyis an integer from the arraynums.1 <= k <= nums.length
Approach 1: Brute Force
Just do what it says.
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;
}
};