Count Number of Maximum Bitwise-OR Subsets | Simplest Approach |Leetcode 2044 | codestorywithMIK
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This is the 116th Video of our Playlist "Array 1D/2D : Popular Interview Problems" by codestorywithMIK
My Subsets Video - • Subsets | Simple Story To Code | Leet...
In this video we will try to solve a medium-easy Array subsets based Problem : Count Number of Maximum Bitwise-OR Subsets | Simplest Approach |Leetcode 2044 | codestorywithMIK
We will solve it using a simple subsets concept.
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 : Count Number of Maximum Bitwise-OR Subsets | Simplest Approach |Leetcode 2044 | codestorywithMIK
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My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Arrays/Count Number of Maximum Bitwise-OR Subsets.cpp
Leetcode Link : https://leetcode.com/problems/count-number-of-maximum-bitwise-or-subsets
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Summary :
Approach-1 (Simple Recursion):
This approach uses recursion to explore all possible subsets of the array. It calculates the bitwise OR of each subset and counts how many subsets have an OR value equal to the maximum OR value of the entire array.
At each step, the function either includes or excludes the current element (nums[idx]), recursively calculates the OR value, and checks if it matches the maximum OR.
Time Complexity: O(2^n) since it explores all subsets.
Space Complexity: O(1) (excluding the recursion stack).
Approach-2 (Memoization):
This is an optimized version of the recursive approach, where we store the results of subproblems (using a memoization table) to avoid recalculating the OR values for the same states.
The table t[idx][currOr] stores the result of subsets from index idx onward with the current OR value currOr. This avoids recomputation and speeds up the process.
Time Complexity: O(n * maxOr) where maxOr is the maximum OR value achievable.
Space Complexity: O(n * maxOr) due to the memoization table.
✨ Timelines✨
00:00 - Introduction
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