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

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Approach

1. First, we have to identify the indexes of key in the nums array, and save into a vector.
2. Iterate through the input vector
3. If the absolute value of iterator i - one of the key indexes is <= k, it satisfies the requirement and should be place in the ouput vector.
4. As soon as one of the key satisfies, stop and proceed to the next index.

Solution

class Solution {
public:
    vector<int> findKDistantIndices(vector<int>& nums, int key, int k) {
        //get index of key
        vector<int> keyIdx;
        vector<int> idx;
        
        for(int i = 0; i < nums.size(); i++){
            if(nums[i] == key) keyIdx.push_back(i);
        }
        
        for(int i = 0; i < nums.size(); i++){
            for(int j = 0; j < keyIdx.size(); j++){
                if(abs(i - keyIdx[j]) <= k){
                    idx.push_back(i);
                    break;
                } 
        }
    }
        return idx;
    }
};