Length of Longest Subarray With at Most K Frequency | 2 Ways | Leetcode 2958 | codestorywithMIK
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This is the 17th Video of our Playlist "Sliding Window : Popular Interview Problems".
In this video we will try to solve an very classic sliding window problem :
Length of Longest Subarray With at Most K Frequency | 2 Ways | Khandani Sliding Window template | Leetcode 2958 | codestorywithMIK
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Length of Longest Subarray With at Most K Frequency | 2 Ways | Khandani Sliding Window template | Leetcode 2958 | codestorywithMIK
Company Tags : will update soon
My solutions on Github(C++ & JAVA) : https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Sliding%20Window/Length%20of%20Longest%20Subarray%20With%20at%20Most%20K%20Frequency.cpp
Leetcode Link : https://leetcode.com/problems/length-of-longest-subarray-with-at-most-k-frequency/
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My Graph Concepts Playlist : https://youtu.be/5JGiZnr6B5w
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Approach Summary :
**Approach-1 (With Nested Loop - Classic Sliding Window Template)**:
This approach utilizes the classic sliding window technique to find the maximum length of a subarray where the number of distinct integers is at most `k`. The algorithm maintains a sliding window represented by two pointers `i` and `j`, where `i` marks the start of the window and `j` marks the end. It uses a hashmap (`unordered_map` in C++) to store the frequency of each integer within the window. The algorithm iterates through the array, expanding the window (`j`) and contracting it (`i`) as necessary to ensure that the number of distinct integers remains at most `k`. The time complexity of this approach is O(n) since it iterates through the array once, and the space complexity is O(n) due to the hashmap storing the frequency of integers.
**Approach-2 (Without Nested Loop)**:
This approach also finds the maximum length of a subarray where the number of occurrences of a particular integer is at most `k`. However, it achieves this without using nested loops. Instead, it maintains a sliding window represented by two pointers `i` and `j`, similar to Approach-1. Additionally, it uses a hashmap (`unordered_map` in C++) to store the frequency of each integer within the window. However, it introduces an additional variable `culprit` to keep track of the number of integers that have exceeded the limit `k` within the window. The algorithm iterates through the array, expanding the window (`j`) and contracting it (`i`) as necessary. If an integer within the window exceeds the limit `k`, it increments `culprit` and adjusts the window accordingly. This approach also has a time complexity of O(n) and a space complexity of O(n).
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✨ Timelines✨
00:00 - Introduction
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