Subarray Product Less Than K | Khandani Sliding Window template | Leetcode 713 | codestorywithMIK
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This is the 16th Video of our Playlist "Sliding Window : Popular Interview Problems".
In this video we will try to solve an very classic sliding window problem :
Subarray Product Less Than K | Khandani Sliding Window template | Leetcode 713 | 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 : Subarray Product Less Than K | Khandani Sliding Window template | Leetcode 713 | codestorywithMIK
Company Tags : Yatra
My solutions on Github(C++ & JAVA) : https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Sliding%20Window/Subarray%20Product%20Less%20Than%20K.cpp
Leetcode Link : https://leetcode.com/problems/subarray-product-less-than-k/
My DP Concepts Playlist : https://youtu.be/7eLMOE1jnls
My Graph Concepts Playlist : https://youtu.be/5JGiZnr6B5w
My Recursion Concepts Playlist : https://www.youtube.com/watch?v=pfb1Zduesi8&list=PLpIkg8OmuX-IBcXsfITH5ql0Lqci1MYPM
My Sliding Window Playlist : https://www.youtube.com/watch?v=mrUBUWb23hk&list=PLpIkg8OmuX-J2Ivo9YdY7bRDstPPTVGvN&index=1
My GitHub Repo for interview preparation : https://github.com/MAZHARMIK/Interview_DS_Algo
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Approach Summary :
This Java code implements a method `numSubarrayProductLessThanK` which calculates the number of subarrays within an integer array `nums` whose product is less than a given integer `k`. The algorithm maintains two pointers (`left` and `right`) to define the current subarray and iterates through the array, adjusting the pointers and calculating the product of the subarray as it goes. If the product exceeds or equals `k`, it moves the `left` pointer forward until the product falls below `k` again. It accumulates the count of such subarrays and returns the total count once the iteration completes. This approach achieves the desired computation efficiently with a time complexity of O(n), where n is the length of the input array `nums`.
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✨ Timelines✨
00:00 - Introduction
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