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Suppose we have a list of numbers called nums and another number k, we have to find a list of count of distinct numbers in each window of size k.

So, if the input is like nums = [2, 2, 3, 3, 4], k = 2, then the output will be [1, 2, 1, 2], as the windows are [2, 2], [2, 3], [3, 3], and [3, 4].

To solve this, we will follow these steps −

c := make a dictionary of elements in nums and their frequencies

ans := a new list

for i in range k to size of nums, do

insert size of c at the end of ans

c[nums[i]] := c[nums[i]] + 1

c[nums[i - k]] := c[nums[i - k]] - 1

if c[nums[i - k]] is same as 0, then

remove c[nums[i - k]]

insert size of c at the end of ans

return ans

Let us see the following implementation to get better understanding −

from collections import Counter class Solution: def solve(self, nums, k): c = Counter() for i in range(k): c[nums[i]] += 1 ans = [] for i in range(k, len(nums)): ans.append(len(c)) c[nums[i]] += 1 c[nums[i - k]] -= 1 if c[nums[i - k]] == 0: del c[nums[i - k]] ans.append(len(c)) return ans ob = Solution() nums = [2, 2, 3, 3, 4] print(ob.solve(nums, 2))

[2, 2, 3, 3, 4], 2

[1, 2, 1, 2]

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