Special Array I - Leetcode 3151 - Python

NeetCodeIO · Intermediate ·⚡ Algorithms & Data Structures ·1y ago

Key Takeaways

The video solves the Leetcode problem Special Array I using Python, iterating over the input array to check if every adjacent pair of elements has a different parity.

Full Transcript

hey everyone welcome back and let's write some more neat code today so today let's solve the problem special array one it always feels really weird after we get done solving some hard problems to be given like an easy problem I'm not going to lie like this one is pretty easy I think for most of us but I know that there might be like some beginners watching this and if you're a beginner it might not necessarily be easy for you but yeah anyways let's get into it so the idea is that we are given an array of elements let me use the second example we want to determine if the input array is is considered special or not we know that the array is special if every adjacent pair of elements so basically like this pair of elements is adjacent this pair of elements is adjacent if we had more elements we would kind of just keep doing that length to Window and what we want to know is that every adjacent pair of elements has a different parody probably the only question in your mind is if you're watching this video what does parody even mean and they kind of explain it I think in one of the other examples but I'll just tell you basically if one of the elements is even then the other one has to be odd we cannot have two consecutive elements that are both odd or that are both even so if this were the case two adjacent odd elements we would return false if we had two adjacent even elements we would return false if we go through the entire input array and neither of those happen to be the case finally at that point we would return true so knowing that we can solve the problem pretty easily just compare every adjacent pair of elements the way I'm going to do this is by having an ey pointer it's not going to start at zero it's actually going to start at one it's going to go up until the end of the array we're always going to compare this element with the previous element and what we're specifically going to do is just take this number mod it by two we're going to get zero take this number mod it by two we're going to get one these are not equal that's good we continue we always expect them to not be equal if they were ever equal then we would return false that's not the case right now we look at the next pair again not equal we keep going another way to think about it I guess would be the binary representation of this number is going to be zero the binary representation of this is going to be this what we really care about is that there are no two adjacent elements that have the same bit set in this position now imagine if I had like an eight over here eight in binary will look like this so in this example neither of them has the bit set so basically they should have a different bit in this position and if we had two even elements that are adjacent at that point we would return false if that didn't exist then we would reach the end of the input array here and then return true so now let's code it up so I'm just going to iterate over the input array nums while starting at index one and then going up until the end of it and what I'm going to check is specifically this the number at index I I'm going to either mod it by two or bit wise and it with one those are both pretty much equivalent and if that's equal to goms of I minus one again I could do this as this way or I could do it this way it'll give me the same thing either way at that point I would return false if that never happens out here I can return true let me run this I'm not 100% sure if I need to wrap this and this in parentheses but let's just run it and try okay it looks like it works it's pretty efficient if you found this helpful check out knode iio for a lot more thanks for watching and I'll see you soon

Original Description

🚀 https://neetcode.io/ - A better way to prepare for Coding Interviews 🧑‍💼 LinkedIn: https://www.linkedin.com/in/navdeep-singh-3aaa14161/ 🐦 Twitter: https://twitter.com/neetcode1 ⭐ BLIND-75 PLAYLIST: https://www.youtube.com/watch?v=KLlXCFG5TnA&list=PLot-Xpze53ldVwtstag2TL4HQhAnC8ATf Problem Link: https://leetcode.com/problems/special-array-i/ 0:00 - Read the problem 0:30 - Drawing Explanation 2:50 - Coding Explanation leetcode 3151 #neetcode #leetcode #python
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The video teaches how to solve the Special Array I problem on Leetcode using Python, by iterating over the input array and checking the parity of adjacent elements. This problem is useful for practicing basic algorithm design and programming fundamentals.

Key Takeaways
  1. Initialize a pointer at index 1 of the input array
  2. Compare the parity of the current element with the previous element
  3. Use the modulus operator or bitwise operations to check for parity
  4. Return false if two adjacent elements have the same parity
  5. Return true if the entire array is iterated over without finding two adjacent elements with the same parity
💡 The key to solving this problem is to understand the concept of parity and how to use bitwise operations or the modulus operator to check for it.

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Chapters (3)

Read the problem
0:30 Drawing Explanation
2:50 Coding Explanation
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