LeetCode 1768: Merge Strings Alternately ·

Table of Contents

Meta Data
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Difficulty: easy First Attempt: 2025-03-18

Submissions: Accepted → Time Limit Exceeded → Accepted

Approach
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Starting with word1, take one character from each string in turn; once one runs out, append the rest of the other.

My Attempts
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The first version used zip until the shorter string ran out, then computed how many characters had been used with round(len(res) / 2) and appended both remainders. Correct, but computing the length with round is roundabout.

The second version used a hand-written index:

idx = 0
while idx < len(word1) or idx < len(word2):
    res += word1[idx] + word2[idx]          # idx never increases

It forgot idx += 1, looping forever and timing out. And with or, it would also read past the end of the shorter string. Using and plus idx += 1 passed.

Solution
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class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        res = []
        i = 0
        while i < len(word1) and i < len(word2):
            res.append(word1[i])
            res.append(word2[i])
            i += 1
        return "".join(res) + word1[i:] + word2[i:]

When the loop ends, at least one string is used up, so word1[i:] + word2[i:] appends only the one with characters left.

Complexity
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Time O(n + m), space O(n + m).

Mistakes to Avoid
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Forgetting to advance the index inside a while loop.
Using or in the condition: it reads past the end of the shorter string.
Concatenating with +=: Python strings are immutable, so each step builds a new string; collect into a list and join.

Interview Explanation
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One index walks both strings, taking a character from each in turn until one runs out. At that point at least one side is exhausted, so I append both remainders. Collecting characters in a list and joining keeps it O(n + m).