Number of Ways to Split Array - Leetcode 2270 - Python

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

Key Takeaways

The video demonstrates a solution to the Leetcode 2270 problem, 'Number of Ways to Split Array', using a prefix sum approach in Python, with a focus on efficient linear time and constant space complexity.

Full Transcript

hey everyone welcome back and let's write some more neat code today so today let's solve a problem number of ways to split array and I'm not going to lie for a beginner like this problem there are some like learnings that you can take away from it I thought it was a little bit on the easy side just in terms of like the acceptance rate I saw the acceptance rate was like 50% and I solved a different prefix sum problem that I thought was much more difficult than this one which had like an 80% completion rate so my takeway is that these completion rates are kind of random so I'm not really going to pay attention to them anymore but to get into this problem we're given an array of integers there actually could be negative integers and that's not really going to change the problem very much at all imagine we're given this as an input array there is something called a split in this problem the definition of it is basically splitting the array into two parts where the left and right parts are both nonm you can't like SP split here this would be empty you can't split here either this would be empty now why didn't they Define the problem like that I don't know but it's not that hard I think to take the way they worded this problem and arrive at this kind of meaning so non-empty splits now if you do perform a split specifically we are interested in the sum of each portion I'll call it the left and right portion so the sum here is 10 the sum here is I believe this will be -1 + 4 I think that's going to be three what we need to determine is this current split valid a valid split is defined as where the left side is a sum of greater than or equal to the right side and this does pass that definition so they gave us that like over here so this is a valid split what we want to do is count the total number of valid splits so we'll do the same exact thing uh rather than here splitting it we're going to split here and we're going to split here so there's only three possible splits that we could do now obviously for an arbitrary split like this one if we were to take this total sum and then this total sum that's an O of n operation for any given split it is like that but once we do a single split can you see how we can then eliminate some of the repeated work if I had this split already and I already had the left and right sum what could I do to then arrive at this split and get me this left and right sum in constant time can I reuse some of this to get this and the answer is yes because obviously the main difference between this line and this line is just this little portion over here if I already have the left sum I can get this left sum just by taking this value and adding it there how do I get the right sum well whatever it already was I can remove this value giving me the new right sum over here and that's how you can solve this problem in linear time and constant space I'll quickly do the dry run the easiest thing to do is just to get the initial sum of all the elements and let's call that our right sum I believe it's going to be 12 that's going to be four I think it's roughly 11 hopefully I'm correct and then now we're going to start iterating our left sum we can initially say is equal to zero and now I'm going to update it by taking the 10 and adding it to the left sum I'm going to use a different color and this will be 10 and I'm going to remove 10 from here that'll be 1 okay I was I was wrong I think the initial sum was 13 I really don't know how my math skills have gotten so poor I thought this was 12 but anyways removing 10 we will get a positive three over here and so right now this is valid this split is valid so we found a valid split okay now let's try the next split let's take four and add it over here we get 14 and remove it over here we get -1 this is also a valid split so we found two valid splits let's try uh the next spot over here let's take 8 and add it over here I think that will give us a postive 6 and let's take 8 and remove it from here that will actually add 8 to this making this a positive 7 and so now the condition doesn't hold this is not greater than or equal to that it makes sense this is 7even by itself and all of this should be postive 6 so we only counted two valid splits at this point we can stop because we don't need to actually be over here because we don't want to do the split so when we iterate over here we will technically only over the first nus1 elements now let's go ahead and code it up so like I said I'm going to initialize my right as just being the sum of all the elements technically this is an O of n operation but we only needed to do that once and now we can iterate over the input like this for I in range length minus one because we don't want to include the last element we can also initialize our left with zero and our result I guess also Z that's the one that we're going to end up returning and so now in here it's just going to be as simple as those steps I talked about we can add to the left the current uh element and from the right we can subtract the current element and then we just need an if statement to determine if we're going to be incrementing the result or not so if the left is greater than or equal to the right we increment the result and that's the solution if you want to shorten this a tiny bit you can use use a Turner operator and you can say something like this add one if left greater than right else add zero and then you can get rid of those two lines even though we have a loop here and this is also an oent operation this is still clearly a linear time solution it's just two passes I guess and there's no extra data structures being used except of course I forgot the equal sign when I was rewriting it into a Turner operator uh but you can see that this is working code it's pretty efficient This falls into the prefix some pattern if you want to learn more about that you can check out NE code. I think I cover it in my Advanced algorithms course 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/number-of-ways-to-split-array/description/ 0:00 - Read the problem 0:30 - Drawing Explanation 5:04 - Coding Explanation leetcode 2270 #neetcode #leetcode #python
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The video teaches how to solve the Leetcode 2270 problem using a prefix sum approach, focusing on efficient time and space complexity. It covers the problem definition, algorithm design, and implementation in Python.

Key Takeaways
  1. Initialize the right sum as the total sum of the array elements
  2. Iterate over the array, updating the left and right sums for each split
  3. Check if the current split is valid and increment the result if it is
  4. Return the total number of valid splits
💡 The prefix sum pattern can be used to solve this problem efficiently by avoiding repeated work and reducing time complexity to linear.

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Chapters (3)

Read the problem
0:30 Drawing Explanation
5:04 Coding Explanation
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