Count Distinct Elements in Every Window || Program 57 || Competitive Coding || Learning Monkey ||
Count Distinct Elements in Every Window
In this class, We discuss Count Distinct Elements in Every Window.
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Question:
Given an array of N elements and a window size K.
Our task is to find the distinct element count in each window.
Example:
N = 7 and K = 4
A = [1, 2, 1, 3, 4, 2, 3]
Output: [3, 4, 4, 3]
The first window is [1, 2, 1, 3].
The distinct elements in the window are 3.
Similarly, the remaining windows count.
Time complexity: O(N)
Space Complexity: O(N)
Logic:
We use a hash table to maintain the distinct elements.
The step-by-step explanation is provided in the video.
Code:
from collections import defaultdict
class Solution:
def countDistinct(self, A, N, K):
mp = defaultdict(lambda: 0)
lis=[]
dist_count = 0
for i in range(K):
if mp[A[i]] == 0:
dist_count += 1
mp[A[i]] += 1
lis.append(dist_count)
for i in range(K, N):
if mp[A[i - K]] == 1:
dist_count -= 1
mp[A[i - K]] -= 1
if mp[A[i]] == 0:
dist_count += 1
mp[A[i]] += 1
lis.append(dist_count)
return lis
A=[1,2,1,3,4,2,3]
ob=Solution()
z=ob.countDistinct(A,7,4)
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