Merge Strings Alternately LeetCode

Key takeaways:
Given two strings word1 and word2, merge them by alternating characters starting with word1. If one string is longer than the other, append the remaining characters to the end.
The constraints are 1 <= word1.length, word2.length <= 100. Both strings consist of lowercase English letters.
Use two pointers, i for word1 and j for word2, both starting at index 0. Append characters alternately from both words to the result string.If one string is exhausted, append the remaining characters of the longer string.
The time complexity is $O(m + n)$ , where m is the length of word1 and n is the length of word2.
The space complexity is $O(1)$ , as the solution uses only constant extra space for variables and pointers.

In this Answer, we’ll explore the Merge Strings Alternatively problem, which is asked in technical interviews at top tech companies such as Google, Facebook, Amazon, Uber, and Microsoft. We will review a few sample examples, provide short quizzes related to the problem to brainstorm an optimized approach to the problem, and discuss efficient approaches to solve it. By the end of this Answer, you’ll have a solid understanding of how to tackle this problem effectively in your interviews.

Problem statement

Given two strings word1 and word2, we have to merge the strings by adding letters in alternating order, starting with word1. If a string is longer than the other, append the additional letters to the end of the merged string. Return the merged string.

Constraints:

$1 \leq$ word1.length, word2.length $\leq 100$
word1 and word2 consist of lowercase English letters.

Merge strings alternately solution

To merge strings alternatively, we need to keep track of the characters in both words. To do this, the most feasible approach that should come to mind is to initialize two pointers, both pointing at the first characters of both words.

Let’s say we have a pointer i, and a pointer j, both initialized to 0 and a results string that will be used to return the final output. We start by iterating over both strings until i or j reach the end of word1 or word2 respectively. We first check if i is less than the length of word1, if this is TRUE, we append this character from word1 to results and increment i. Then we check the same thing for word2, if j is less than the length of word2, we append this character from word2 to results and increment j. We keep repeating this process while there are remaining characters in either word1 or word2. Once all of the characters have been appended to results, we return.

Let’s look at the following illustration to get a better understanding of the solution:

def merge_strings_alternately(word1, word2):
    m = len(word1)
    n = len(word2)
    i = 0
    j = 0
    result = ""
    # Iterate through both strings
    while i < m or j < n:
        if i < m:
            result += word1[i] # Append character from word1
            i += 1
        if j < n:
            result += word2[j] # Append character from word2
            j += 1
    return result
def main():
    word1_list = ["abc", "astrology", "universe"]
    word2_list = ["def", "physics", "galaxy"]
    for i in range(len(word1_list)):
        print(f"{i + 1}.\t word1 = '{word1_list[i]}'")
        print(f"\t word2 = '{word2_list[i]}'")
        result = merge_strings_alternately(word1_list[i], word2_list[i])
        print(f"\n\t After merging, the final string becomes: '{result}'")
        print("-" * 100)
if __name__ == "__main__":
    main()

Merge Strings Alternatively

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Merge Strings Alternately LeetCode

Problem statement

Knowledge test

Merge strings alternately solution

Coding example

Time complexity

Space complexity