Maximum Width Ramp | Brute Force | Better | Optimal | Leetcode 962 | codestorywithMIK
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This is the 8th Video of our Playlist "Two Pointers : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a good 2 Pointer based Problem : Maximum Width Ramp | Brute Force | Better | Optimal | Leetcode 962 | codestorywithMIK
NOTE : This problem is an ideal candidate for "Monotonic Stack" but as you all know, there is some medical urgency at my home because of which I couldn't get much time today. Hence posting 2 Pointer approach only. But as soon as I get time, I am going to post a detailed explanation on Monotonic Stack approach.
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 | Brute Force | Better | Optimal | Leetcode 962 | codestorywithMIK
Company Tags : Google, Amazon
My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Arrays/Two Pointer/Maximum Width Ramp.cpp
Leetcode Link : https://leetcode.com/problems/maximum-width-ramp
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Summary :
Approach 1: Brute Force
Description: This approach uses a nested loop to check every possible pair (i, j) where i less than j and nums[i] less than equal nums[j]. If a valid pair is found, it updates the maximum ramp width (j - i).
Time Complexity: O(n2), since it involves two loops iterating through the array.
Space Complexity: O(1), as no additional data structures are used.
Limitations: Inefficient for large inputs and fails some test cases due to time limits.
Test Case Success: Passes 95/101 test cases.
Approach 2: Early Termination
Description: Similar to the brute-force approach but optimizes by starting the inner loop from the end (j = n - 1) and breaking early once a valid ramp is found for a given i. This avoids unnecessary comparisons for each i.
Time Complexity: O(n2) in the worst case, but performs better in practice due to early termination.
Space Complexity: O(1), as it does not require extra space.
Limitations: Somewhat more efficient than Approach 1 but still can fail on large inputs.
Test Case Success: Passes 97/101 test cases.
Approach 3: Two-Pointer with Preprocessing (maxRight Array)
Description: This approach preprocesses the array by creating a maxRight array, where maxRight[i] stores the maximum element from index i to the end. Then, two pointers (i and j) are used to find the maximum width ramp. If nums[i] is less than or equal to maxRight[j], update ramp and move the j pointer right. Otherwise, increment i.
Time Complexity: O(n), as it involves a linear scan to build maxRight and another linear scan using two pointers.
Space Complexity: O(n), because of the additional maxRight array.
Advantages: Efficient and handles all test cases.
Test Case Success: Passes all test cases and is the Accepted solution.
Overall, Approach 3 is the optimal solution due to its linear time complexity and ability to handle large inputs without exceeding time limits.
✨ Timelines✨
00:00 - Introduction
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