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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 <= 10^5
  • 1 <= 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;
    }
}

  • // OJ: https://leetcode.com/problems/distinct-numbers-in-each-subarray/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> distinctNumbers(vector<int>& A, int k) {
        unordered_map<int, int> cnt;
        vector<int> ans(A.size() - k + 1);
        for (int i = 0, N = A.size(); i < N; ++i) {
            cnt[A[i]]++;
            if (i - k >= 0 && --cnt[A[i - k]] == 0) cnt.erase(A[i - k]);
            if (i >= k - 1) ans[i - k + 1] = cnt.size();
        }
        return ans;
    }
};

  • class Solution:
    def distinctNumbers(self, nums: List[int], k: int) -> List[int]:
        n = len(nums)
        cnt = Counter(nums[:k])
        ans = [len(cnt)]
        for i in range(k, n):
            u = nums[i - k]
            cnt[u] -= 1
            if cnt[u] == 0:
                cnt.pop(u)

            cnt[nums[i]] += 1
            ans.append(len(cnt))
        return ans

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