Find all K-distance indices in array.
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
class Solution {
public List<Integer> findKDistantIndices(int[] nums, int key, int k) {
List<Integer> indices = new ArrayList<Integer>();
for (int i = 0; i < nums.length; i++){
if (nums[i] == key){
for (int j = i-k; j<=i+k; j++){
if (j >= 0 && j < nums.length; j++) {
if (!indices.contains(j)) {
indices.add(j)
}
}
}
}
}
return indices;
}
}
10, 1, 1 => [0,1]
101,1,1 =>[0,1,2]
1010001,1,1 => [0,1,2,3,5,6]
1010001,1,2 => [0,1,2,3,4,5,6]
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