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Count distinct elements in every window | #57 - GFG 160-Day Challenge| GFG POTD

In this video, we tackle the GeeksforGeeks Problem of the Day (POTD): Count Distinct Elements in Every Window, as part of the GFG 160-Day Challenge. This problem involves efficient use of sliding window and hashing techniques to find the count of distinct elements in all subarrays of size K.
🔍 What You'll Learn in This Video:

1. Problem Explanation:
Understand the problem requirements and constraints for calculating distinct elements in every subarray of size K.

2. Optimal Approach Using Hash Maps:
Learn how to use a sliding window and hash map for an efficient O(n) solution.
3. Time Complexity Optimization:
Avoid the naive O(n*k) approach by leveraging efficient data structures.
💡 Problem Statement Highlights:

Given an array of integers and a window size K, count the number of distinct elements in every contiguous subarray (window) of size K.
📚 Example:

Input:
arr = [1, 2, 1, 3, 4, 2, 3]
K = 4

Output:
3 4 4 3

Explanation:

Window [1, 2, 1, 3] has 3 distinct elements: 1, 2, 3.

Window [2, 1, 3, 4] has 4 distinct elements: 1, 2, 3, 4.

Window [1, 3, 4, 2] has 4 distinct elements: 1, 2, 3, 4.

Window [3, 4, 2, 3] has 3 distinct elements: 2, 3, 4.
📌 Useful Links:

Problem Link :- https://www.geeksforgeeks.org/batch/gfg-160-problems/track/two-pointer-technique-gfg-160/problem/count-distinct-elements-in-every-window

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