LeetCode 1768: Merge Strings Alternately — Python Solution

LeetCode 1768, Merge Strings Alternately, is a linear traversal problem: emit one character from each input while both have characters left, then append the unused suffix from the longer string.

Clear Python solution

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        merged = []
        limit = min(len(word1), len(word2))

        for index in range(limit):
            merged.append(word1[index])
            merged.append(word2[index])

        merged.append(word1[limit:])
        merged.append(word2[limit:])
        return "".join(merged)

Why this handles unequal lengths

The loop stops at the shorter input. At that point, exactly one suffix can be non-empty, but appending both slices is simpler and correct because appending an empty string has no effect. This avoids exception-driven control flow and makes the boundary obvious.

Complexity

The algorithm reads each input character once and creates one output character per input character: O(m + n) time and O(m + n) output space. A list plus join avoids repeatedly reallocating immutable Python strings.

Example

For word1 = "ab" and word2 = "pqrs", the loop emits apbq, then the remaining suffix is rs, producing apbqrs.

Problem reference: LeetCode 1768: Merge Strings Alternately.