1768.Merge_Strings_Alternately(Easy).md

1768. Merge Strings Alternately (Easy)

Date and Time: Sep 13, 2024, 11:23 (EST)

Link: https://leetcode.com/problems/merge-strings-alternately/

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.

Walk-through:

Similar to how to merge two linked list, we first add chars alternately when both i, j for word1 and word2 have the same length, then, if i < len(word1) or j < len(word2), we add all the rest into res.

Python Solution:

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        # i, j ptrs for word1, word2, when both are in bound, add char alternatively
        # Then, check if i or j has remaining chars to append

        # TC: O(n+m), n=len(word1), m=len(word2), SC: O(n+m)
        i, j = 0, 0
        res = ""
        while i < len(word1) and j < len(word2):
            res += word1[i] + word2[j]
            i, j = i + 1, j + 1
        # Add remaining chars
        if i < len(word1) or j < len(word2):
            res += word1[i:] or word2[j:]
        return res

Time Complexity: $O(n + m)$, n is the length of word1, m is the length of word2.
Space Complexity: $O(n+ m)$