README_EN.md

1852. Distinct Numbers in Each Subarray

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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

Python3

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

Java

class Solution {
    public int[] distinctNumbers(int[] nums, int k) {
        int[] cnt = new int[100010];
        int x = 0;
        for (int i = 0; i < k; ++i) {
            if (cnt[nums[i]]++ == 0) {
                ++x;
            }
        }
        int n = nums.length;
        int[] ans = new int[n - k + 1];
        ans[0] = x;
        for (int i = k; i < n; ++i) {
            if (--cnt[nums[i - k]] == 0) {
                --x;
            }
            if (cnt[nums[i]]++ == 0) {
                ++x;
            }
            ans[i - k + 1] = x;
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> distinctNumbers(vector<int>& nums, int k) {
        int cnt[100010] = {0};
        int x = 0;
        for (int i = 0; i < k; ++i) {
            if (cnt[nums[i]]++ == 0) {
                ++x;
            }
        }
        int n = nums.size();
        vector<int> ans(n - k + 1);
        ans[0] = x;
        for (int i = k; i < n; ++i) {
            if (--cnt[nums[i - k]] == 0) {
                --x;
            }
            if (cnt[nums[i]]++ == 0) {
                ++x;
            }
            ans[i - k + 1] = x;
        }
        return ans;
    }
};

Go

func distinctNumbers(nums []int, k int) []int {
	cnt := map[int]int{}
	for _, v := range nums[:k] {
		cnt[v]++
	}
	ans := []int{len(cnt)}
	for i := k; i < len(nums); i++ {
		u := nums[i-k]
		cnt[u]--
		if cnt[u] == 0 {
			delete(cnt, u)
		}
		cnt[nums[i]]++
		ans = append(ans, len(cnt))
	}
	return ans
}

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