Formatted question description: https://leetcode.ca/all/1852.html
1852. Distinct Numbers in Each Subarray
Level
Medium
Description
Given an integer array nums and an integer k, you are asked to construct the array ans of size n-k+1 where ans[i] is the number of distinct numbers in the subarray nums[i:i+k-1] = [nums[i], nums[i+1], ..., nums[i+k-1]].
Return the array ans.
Example 1:
Input: nums = [1,2,3,2,2,1,3], k = 3
Output: [3,2,2,2,3]
Explanation: The number of distinct elements in each subarray goes as follows:
- nums[0:2] = [1,2,3] so ans[0] = 3
- nums[1:3] = [2,3,2] so ans[1] = 2
- nums[2:4] = [3,2,2] so ans[2] = 2
- nums[3:5] = [2,2,1] so ans[3] = 2
- nums[4:6] = [2,1,3] so ans[4] = 3
Example 2:
Input: nums = [1,1,1,1,2,3,4], k = 4
Output: [1,2,3,4]
Explanation: The number of distinct elements in each subarray goes as follows:
- nums[0:3] = [1,1,1,1] so ans[0] = 1
- nums[1:4] = [1,1,1,2] so ans[1] = 2
- nums[2:5] = [1,1,2,3] so ans[2] = 3
- nums[3:6] = [1,2,3,4] so ans[3] = 4
Constraints:
1 <= k <= nums.length <= 10^51 <= nums[i] <= 10^5
Solution
Use a hash map to store each number’s occurrences in the sliding window. For each i from k - 1 to n - 1, the corresponding value in ans is the size of the map.
class Solution {
public int[] distinctNumbers(int[] nums, int k) {
int n = nums.length;
int[] ans = new int[n - k + 1];
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int i = 0; i < k; i++) {
int num = nums[i];
map.put(num, map.getOrDefault(num, 0) + 1);
}
ans[0] = map.size();
for (int i = k; i < n; i++) {
int prev = nums[i - k], curr = nums[i];
map.put(prev, map.get(prev) - 1);
if (map.get(prev) == 0)
map.remove(prev);
map.put(curr, map.getOrDefault(curr, 0) + 1);
ans[i - k + 1] = map.size();
}
return ans;
}
}