Shortest Subarray With OR at Least K II | Made Easy | Detailed | Leetcode 3097 | codestorywithMIK
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This is the 23rd Video of our Playlist "Bit-Manipulation : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a good bit-manipulation problem : Shortest Subarray With OR at Least K II | Made Easy | Detailed | Leetcode 3097 | 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 : Shortest Subarray With OR at Least K II | Made Easy | Detailed | Leetcode 3097 | codestorywithMIK
Company Tags : will update
My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Bit_Magic/Shortest Subarray With OR at Least K II.cpp
Leetcode Link : https://leetcode.com/problems/shortest-subarray-with-or-at-least-k-ii
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Summary :
Bitwise Representation Tracking: We use a vec array of size 32 to keep track of the number of set bits (1s) at each bit position across elements in the current window. This allows us to dynamically calculate the bitwise OR of all elements in the window without directly recomputing it each time.
Sliding Window: We maintain a sliding window with two pointers, i (start) and j (end). We expand the window by moving j and adding nums[j] to vec.
Window Contraction: If the current OR of the window (determined from vec) meets or exceeds the target k, we check if the current window length is the minimum found so far. We then contract the window from the left by moving i and removing nums[i] from vec.
Result: The smallest window length meeting the condition is stored in result, or we return -1 if no valid window is found.
This approach efficiently computes the minimum subarray length with a bitwise OR greater than or equal to k by leveraging bit manipulation and a sliding window.
✨ Timelines✨
00:00 - Introduction
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