Merge Strings Alternately - Leetcode 1768 - Python

NeetCodeIO · Intermediate ·⚡ Algorithms & Data Structures ·3y ago

Key Takeaways

The video demonstrates a solution to the Leetcode problem 1768, Merge Strings Alternately, using Python. The problem involves merging two strings character by character, starting with the first character from the first string, then the first character from the second string, and so on.

Full Transcript

hey everyone welcome back and let's write some more neat code today so today let's solve the problem merge strings alternately we're given two strings word one and word two and basically we want to merge them character by character starting with the first character from word one and then adding that to the output string so for example if we have two words one word one is ABC and word two is p q r we're going to start with the first character from word one and add that to the output and then we're gonna get the first character of word two which is p add that to the output and then get the next character from word one which is B and the next character from word two q c here and then R from this word and then this is the output string and that's what we would return just like we have over here now in this example both strings were of equal size but we could run into a case where one of the words is much longer than the other one so for example if we got to the point where we were at this where we took the first three characters from each string and then this is what we had and then we run out of characters from word one nothing left here but we have a bunch of characters in word two well then in that case we would just take whatever is over here let's say it's x y z something something something and then just append that to the string that we currently have so we were just have x y z and whatever else comes over here so algorithmically the way we're going to code this up is just by having a pointer for word one let's say that's I and have a pointer for word two let's say that's J and just keep iterating through these while both of them are in bounds and when one of them goes out of bounds we take the all the characters from the other string and append that to the result that we had so overall the algorithm is going to run an O of n plus M time because word one let's say has n characters in word two maybe has M characters we have to visit every single character from each string so now let's code it up so I'm going to start with initializing our pointers I and J I'm going to set both of them equal to zero they're going to be at the beginning of each of the words and we're going to be building our result and even though the result is going to be a string a pending and updating a string is not an efficient operation it's more efficient to just add the strings as as characters to an array and then at the end we will join all of the characters in that array like this in Python you can use that empty string as the delimiter and basically say all the strings and result are going to be appended together with this in between them and this is an empty string so basically nothing in between them so now let's actually build that output string well first of all we're going to be iterating while the eye pointer is in bounds while it is in bounds of the first word and while J is in bounds of the second word as that's the case we're going to take whatever character is at index I in word one and append that to our result like this and I'm going to copy and paste this line because that's what we're basically going to do with word two as well except with pointer J so just like that and then don't forget to actually update the pointer so I is going to be incremented by 1 J is all also going to be incremented by one at the end of this one of the strings could still have characters left so in Python the easiest thing to do here would just be to append to the result well append like this since result is an array the substring of word one starting at index I going up until the end of the string so if I is out of bounds this will append nothing to result but if I is in bounds we'll append the remaining characters of word one and we're going to do the same thing with word two because we don't know which one of them could still have characters left so with bird two we're going to use pointer J and just like that and this will ultimately return the result we want so let's run it to make sure that it works as you can see yes it does and it's pretty efficient if this was helpful please like And subscribe if you're preparing for coding interviews check out neat code.io it has a ton of free resources to help you prepare thanks for watching and hopefully I'll see you pretty soon

Original Description

🚀 https://neetcode.io/ - A better way to prepare for Coding Interviews 🥷 Discord: https://discord.gg/ddjKRXPqtk 🐦 Twitter: https://twitter.com/neetcode1 🐮 Support the channel: https://www.patreon.com/NEETcode ⭐ BLIND-75 PLAYLIST: https://www.youtube.com/watch?v=KLlXCFG5TnA&list=PLot-Xpze53ldVwtstag2TL4HQhAnC8ATf 💡 DYNAMIC PROGRAMMING PLAYLIST: https://www.youtube.com/watch?v=73r3KWiEvyk&list=PLot-Xpze53lcvx_tjrr_m2lgD2NsRHlNO&index=1 Problem Link: https://neetcode.io/problems/merge-strings-alternately 0:00 - Read the problem 0:30 - Drawing Explanation 2:00 - Coding Explanation leetcode 1768 #neetcode #leetcode #python
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This video teaches how to solve the Leetcode problem 1768, Merge Strings Alternately, using Python. The solution involves using two pointers to iterate through the strings and merge them character by character.

Key Takeaways
  1. Initialize two pointers, one for each string
  2. Iterate through the strings using the pointers
  3. Merge the characters from each string
  4. Append the remaining characters from the longer string
  5. Return the merged string
💡 Using two pointers to iterate through the strings allows for an efficient solution with a time complexity of O(n + m)

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Chapters (3)

Read the problem
0:30 Drawing Explanation
2:00 Coding Explanation
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