1852. Distinct Numbers in Each Subarray π
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 <= 105
1 <= nums[i] <= 105
Solutions
Solution 1: Sliding Window + Hash Table
We use a hash table $cnt$ to record the occurrence times of each number in the subarray of length $k$.
Next, we first traverse the first $k$ elements of the array, record the occurrence times of each element, and after the traversal, we take the size of the hash table as the first element of the answer array.
Then, we continue to traverse the array from the index $k$. Each time we traverse, we increase the occurrence times of the current element by one, and decrease the occurrence times of the element on the left of the current element by one. If the occurrence times of the left element become $0$ after subtraction, we remove it from the hash table. Then we take the size of the hash table as the next element of the answer array, and continue to traverse.
After the traversal, we return the answer array.
The time complexity is $O(n)$, and the space complexity is $O(k)$. Where $n$ is the length of the array $nums$, and $k$ is the parameter given by the problem.
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