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Question

You are given two strings word1 and word2. Merge the strings by adding letters in alternating order, starting with word1. If a string is longer than the other, append the additional letters onto the end of the merged string.

Return the merged string.

Example 1:
Input: word1 = "abc", word2 = "pqr"
Output: "apbqcr"
Explanation: The merged string will be merged as so:
word1: a b c
word2: p q r
merged: a p b q c r

Example 2:
Input: word1 = "ab", word2 = "pqrs"
Output: "apbqrs"
Explanation: Notice that as word2 is longer, "rs" is appended to the end.
word1: a b
word2: p q r s
merged: a p b q r s

Example 3:
Input: word1 = "abcd", word2 = "pq"
Output: "apbqcd"
Explanation: Notice that as word1 is longer, "cd" is appended to the end.
word1: a b c d
word2: p q
merged: a p b q c d

Constraints:
1 <= word1.length, word2.length <= 100
word1 and word2 consist of lowercase English letters.

My Solutioin

Code

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        result = ''
        if len(word1) < len(word2):
            shorter_word = word1
            longer_word = word2
        else:
            shorter_word = word2
            longer_word = word1

        shorter_len = len(shorter_word)
        for i in range(shorter_len):
            result += word1[i] + word2[i]
        result += longer_word[shorter_len:]

        return result

I tried to find the way that have less number of iteration.
Also think of way to avoid "index out of range".
In that process, I think of finding shorter word first.

Feedback

Better Solution?

Fastest Sol.

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

They skipped process of finding shorter word!

Comment

I should know what algorithm is more efficient🤔