Maximum Width Ramp | How to identify Monotonic Stack | Dry Runs | Leetcode 962 | codestorywithMIK

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This is the 27th Video of our Playlist "Stack : Popular Interview Problems" by codestorywithMIK

In this video we will try to solve a very good Monotonic Stack Problem : Maximum Width Ramp | How to identify Monotonic Stack | Dry Runs | Leetcode 962 | codestorywithMIK
We have already solved this problem using simple 2 pointer approach -    • Maximum Width Ramp | Brute Force | Better ...  

NOTE : At the timestamp 2:33, I have focused on "How to identify if a problem can be solved by Monotonic Stack".

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 : Maximum Width Ramp | How to identify Monotonic Stack | Dry Runs | Leetcode 962 | codestorywithMIK
Company Tags : Google, Amazon
My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Intervie...
Leetcode Link : https://leetcode.com/problems/maximum...


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Summary :
The maxWidthRamp function uses a monotonic stack approach to find the maximum width ramp (i.e., the largest distance j−i such that nums[i] less than equal to nums[j]) in an array. The solution follows these steps:

Stack Construction: Traverse the array from left to right, pushing indices onto a stack if they represent a decreasing subsequence. This stack will store potential left boundaries for ramps, maintaining indices in a way that each new index has a smaller or equal value than previous entries.

Right-to-Left Traversal: Starting from the end of the array, check each index as a potential right boundary. For each index, pop indices from the stack while the values are less than or equal to the current element. Calculate the width j−i with each pop, updating the maximum ramp width found.

This approach efficiently identifies the maximum width by leveraging the stack to quickly find feasible left boundaries for each right boundary.



✨ Timelines✨
00:00 - Introduction

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