Covert 1D Array Into 2D Array - Leetcode 2022 - Python

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

Key Takeaways

The video demonstrates how to convert a 1D array into a 2D array using Python, with a focus on solving the LeetCode problem. The solution involves checking if the length of the input array can be rearranged into a rectangle of given dimensions, and then using a loop to build the 2D array row by row.

Full Transcript

hey everyone welcome back and let's write some more neat code today so today I'll solve the problem convert 1D array into 2D array so it's pretty much the opposite of flattening a multi-dimensional array so a picture is definitely worth a th words here so the idea is that we're given a one-dimensional array it could be of any sort of length so let's call that a length x we want to break it into a rectangle it has to be a rectangle like we can't have a piece of it missing like you know something like this like we can't have this guy missing so it has to be a rectangle and the dimensions of that rectangle are going to be M by n where these are actually parameters that are given to us so we have to take this rearrange all the elements so that they look like this where the first two elements go in the first row the second two elements go in the second row now imagine if we had five elements it's not really possible then to form a rectangle especially with these Dimensions so what we do well in that case we return an empty array as like the default value if we can form the rectangle then we form it and then we return that so those are the two choices that are given to us now you kind of just tell me by looking at it is there a quick rule of thumb or quick formula you could use to determine if x elements can fit inside a grid that looks like this well literally just check if x which remember that's the length of the input is equal to n * m if this is not the case then we return that empty array that we were talking about the default value otherwise we actually form the rectangle cuz if this is not like true if this is actually equal then it must be possible to break it up into rows and columns because these are factors of that number anyways I won't get into the math of that but how exactly do we do it well let's kind of pay attention to the variables we're actually going to be needing so we're going to mainly be needing n and M the length of this doesn't really matter because we've already confirmed this can be rearranged into a rectangle of these Dimensions so what we do is we're going to go row by row I think it makes sense to build this thing row by row so how do we know what's the first row well we start at index zero and we go up until the length of a particular row M tells us the number of rows that we have n tells us the number of columns which tells us the length of an actual row so we're going to go from 0 to nus1 that's the first row and then the next row is going to start at n so this is the second row the third row is going to start at 2 * n the fourth row is going to start at 3 * n etc etc we're just going to keep multiplying that to tell us the beginning of the row so that's easy right now we know the beginning of every single row that's going to go here first it's zero then it's going to be n then it's going to be 2 N then it's going to be 3 n etc etc this tells us the beginning right this tells us the start if I have the start how do I get the end well I know the length of a row is going to be n so the end of this row should be 3 * n + N - 1 so this is the part that we added N - one so this is the beginning Index this is the ending index if this one is actually being included in Python it would not be included so we'd get rid of the minus one but anyways that's the main idea of solving this problem the rest will probably make more sense in the code so I'm going to start with that if statement I was talking about let's just get the length of the original which is a list provided to us if that is not equal to M * n then we can return immediately otherwise we're going to build the 2D array and then finally return it and we can do that like this row by row so I'm going to say for I could say I or r i guess I'll do R just cuz it's a bit more descriptive but this tells us the current row that we're in for R in range M I guess this is something I didn't talk about in the drawing but how many rows do we have well that's this many M rows n tells us this length of a row M tells us how many rows we have so we're going to build this row by row so how do we get the current portion of the array well remember we were thinking of it in terms of pointers the start index and the end index the start is pretty simple it's always going to be n times the current row that we're at that's going to be R time n so if we're in the first row this is going to be zero if we're in the second row this is going to be n then it's going to be 2 N Etc end is going to be pretty easy to calculate after we have start it's just this plus n minus one but I'm actually not going to have this minus one because in Python we can actually do this we can take the original list and get a slice of it from the start index to the end index but it doesn't necessarily include this and for that reason we get rid of the minus one here and then this slice this tells us the row that we want to add to the result so let's just do this that's it let's run this you can see it works and it's pretty efficient I think in terms of Big O time complexity it is going to be the dimensions of the input n * m I'm assuming that those are the dimensions of this it may not necessarily be I'm reusing these from here you can imagine that these are different I guess but a space that's actually interesting we probably could have optimized the space a bit more we didn't need to necessarily take a slice in here we could have had another nested Loop that would go something like this um I don't know for I in range from the start to the end and then appended each individual element to the result if you want to you can do that I'll leave that as an exercise for the reader I encourage you to try that out but I'm not going to do that just because in a real interview I could do that in 2 seconds we all know how to do that but I prefer this solution I don't think the extra space that is is Created from a slice is really going to matter so I'm fine with this so space complexity though of this current solution if we're not counting the result it's going to be the length of one of these rows which we established is n so let's say the space complexity is n if you found this helpful check out n code. 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/convert-1d-array-into-2d-array/description/ 0:00 - Read the problem 0:30 - Drawing Explanation 3:38 - Coding Explanation leetcode 2022 #neetcode #leetcode #python
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This video teaches how to convert a 1D array into a 2D array using Python, with a focus on solving the LeetCode problem. The solution involves checking if the length of the input array can be rearranged into a rectangle of given dimensions, and then using a loop to build the 2D array row by row.

Key Takeaways
  1. Check if the length of the input array can be rearranged into a rectangle of given dimensions
  2. Use a loop to build the 2D array row by row
  3. Use slicing to extract rows from the input array
💡 The key insight is to check if the length of the input array can be rearranged into a rectangle of given dimensions, and then use a loop to build the 2D array row by row.

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Chapters (3)

Read the problem
0:30 Drawing Explanation
3:38 Coding Explanation
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