Merge Strings Alternately - Leetcode 1768 - Python

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

leetcode 1768

#neetcode #leetcode #python

[def__init]

Was asked a LC hard in an interview today that you covered.
Hoping the rest of the rounds go well, so thankful for the content.

[helloworldcsofficial]

This was a simple and easy to understand solution. Thanks!

[jagannthaja]

Would love and appreciate a video on all the basics concepts of data structures and algorithms and there uses. Would help lot of non cs students 🙏🙏
Edited: I think u already hav such videos like big O for coding interviews
Need that kind of videos for DSA

[abdullahqureshi]

This was my solution

class Solution:
def mergeAlternately(self, word1: str, word2: str) -> str:
n1 = 0
n2 = 0
res = ''

while n1 < len(word1) or n2 < len(word2):

if n1 > len(word2) - 1:
res += word1[n1]
n1 += 1
continue
if n2 > len(word1) - 1:
res += word2[n2]
n2 += 1
continue

res += word1[n1] + word2[n2]
n1 += 1
n2 += 1

return res

[m.kamalali]

Great video as always,
I think One pointer will do the job we don't need two

[shubhambiswas]

go for the word with shorter length to loop and then concat what is left on the longer word

[phongtranquoc]

another approach with time complexity O(m+n): "class Solution {
public String mergeAlternately(String word1, String word2) {
String ans = "";
int length1 = word1.length();
int length2 = word2.length();
int index = 0;
for(int i = 0; i < length1; i++) {
ans += word1.charAt(i);
if(i == index && i <= length1 - 1 && index <= length2 - 1) {
ans += word2.charAt(index);
index++;
}
}

if(length1 < length2) {
ans += word2.substring(index, length2);
}
return ans;
}
}"

[NeoGame]

Hi NeetCode, could you please explain why did you assess the time complexity as O(n + m), and not as O(Max(n, m). Because basically the algorithm will loop at max Math.Max(n, m) if we are using while with two conditions.

[ibnjay]

class Solution:
def mergeAlternately(self, word1: str, word2: str) -> str:
output = ""
len1 = len(word1)
len2 = len(word2)
pointer1 = 0
pointer2 = 0
for i in range(len1+len2):
if pointer1 < len1:
output = output + word1[pointer1]
pointer1 = pointer1+1
if pointer2 < len2:
output = output + word2[pointer2]
pointer2 = pointer2 +1
return output

I did it such an unpythonic way

[bort-oyux]

i solved it with a stringbuilder in java idk if that's a correct approach

[RockuHD]

Try pre sizing the list. Bet that speeds it up a fair bit. Also one index.

Better yet use c++, malloc the size of memory you need(+1 for the null terminator), the set the char at each memory address and return a const char* that points to the start of the string

[marwanahmed]

Hey Neetcode, Can you please solve this problem 662. Maximum Width of Binary Tree?

[redone]

Why are you using two i and j counters when they are exactly the same? j can be removed completely and be replaced by i

[Nisha.......]

Leetcode problem number 2096 please!!!

[adhivp]

Can anybody explain why converting string to list and then converting list to string is more efficient in python.

[infinitygod]

I have a question. Do Google know it is you(from your voice and such)