Leetcode 75 #1

1768. Merge Strings Alternately

I have decided to start up a daily/weekly coding challenge for myself on the Leetcode platform to revamp some of my algorithms knowledge I may have forgotten over the years.

Here is the first of the Leetcode 75!

Algorithm	Runtime	Memory
A	32ms	16.6mb
B	25ms	16.5mb

Algorithm A: Two pointers

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        merged_string = ""

        left = right = 0
        while left < len(word1) or right < len(word2):
            if left < len(word1):
                merged_string += word1[left]
                left += 1
            if right < len(word2):
                merged_string += word2[right]
                right += 1
                
        return merged_string

Algorithm B: One Pointer

I rewrote it to just play around and see if converting the original word1 parameter into a list would be any faster. I think it has to do with how python can sometimes utilize the same underlying data structure, and remaps a subset of the old object to create the new object.

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        word1 = list(word1)
        left = 0

        while left < len(word1):
            if left >= len(word2):
                break
            word1.insert(2 * left + 1, word2[left])
            left += 1

        word1 += word2[left:]
                
        return "".join(word1)

Over multiple tests, on average, one pointer was significantly faster (21.00+ %)