Find the Power of K-Size Subarrays I | Simple Explanation | Dry Run| Leetcode 3254 |codestorywithMIK
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In this video we will try to solve a classic Sliding Window Problem : Find the Power of K-Size Subarrays I | Simple Explanation | Clean Dry Run | Leetcode 3254 | 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 : Find the Power of K-Size Subarrays I | Simple Explanation | Clean Dry Run | Leetcode 3254 | codestorywithMIK
Company Tags : will update later
My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Sliding Window/Find the Power of K-Size Subarrays I.cpp
Leetcode Link : https://leetcode.com/problems/find-the-power-of-k-size-subarrays-i
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Summary :
The solution identifies the last element of every subarray of size k that contains strictly consecutive integers in an input array nums. It uses a sliding window technique:
Initialization:
Create a result array of size (n - k + 1) initialized with -1.
Use a count variable to track the length of the current sequence of consecutive integers.
Preprocess the First Window:
Iterate through the first k elements and check if each element is one more than the previous one.
Update count accordingly and mark the result if the window contains exactly k consecutive elements.
Sliding Window:
Slide the window by advancing pointers i (start) and j (end).
For each new element, check if it continues the consecutive sequence.
Update the result array at position i if the sequence meets the condition.
Return the Result:
Return the result array, where valid positions have the last element of a consecutive sequence, and invalid ones remain -1.
✨ Timelines✨
00:00 - Introduction
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