« back

Find All K-Distant Indices in an Array

Tags : leetcode, sort, array, cpp, medium

Problem Statement - link #

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.

Examples #

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 #

Solutions #


class Solution {
public:
    vector<int> findKDistantIndices(vector<int>& nums, int key, int k) {
        int nk = 2*k + 1;
        int n = nums.size();
        vector<int> res;
        int keys = 0;
        for(int i = 0; i < k + 1 && i < n; i++) {
            if(nums[i] == key) {
                keys++;
            }
        }
        if(keys > 0) {
            res.push_back(0);
        }
        for(int i = 1; i < n; i++) {
            if((i - k - 1 >= 0) && nums[i-k-1] == key) {
                keys--;
            }
            if((i + k < n) && nums[i+k] == key) {
                keys++;
            }
            if(keys > 0) {
                res.push_back(i);
            }
        }
        sort(res.begin(), res.end());
        return res;
    }
};