How do you solve Merge Strings Alternately on LeetCode?

Merge Strings Alternately (LeetCode #1768) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using two pointers allows for efficient traversal of both strings.

What is the brute force solution for Merge Strings Alternately?

The brute force approach for Merge Strings Alternately is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves manually merging the two strings by iterating through each character of both strings. We alternate between the two strings until we reach the end of one, then append the remaining characters of the longer string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Merge Strings Alternately?

Merge Strings Alternately uses the following concepts: Two Pointers, String. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for Merge Strings Alternately?

Clarify the problem requirements and constraints before jumping into coding. Discuss your thought process and approach with the interviewer to ensure you're on the right track. Consider edge cases, such as empty strings or very different lengths, during your explanation.

What are common mistakes when solving Merge Strings Alternately?

Forgetting to check if one string is longer than the other and not appending the remaining characters. Not using a list to build the result, leading to inefficient string concatenations.

#1768

Merge Strings Alternately

Easy

Two PointersStringTwo PointersString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses two pointers to efficiently merge the strings in a single pass. This approach minimizes the number of concatenations by building the result in a list and then joining it at the end, which is more efficient than repeatedly concatenating strings.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty list to hold characters.
2Step 2: Use two pointers to iterate through both strings.
3Step 3: For each pointer, append the character from word1 and then from word2 to the list.
4Step 4: After the loop, append any remaining characters from either string to the list.
5Step 5: Join the list into a single string and return it.

solution.py10 lines

1def mergeAlternately(word1, word2):
2    result = []
3    i = 0
4    while i < len(word1) or i < len(word2):
5        if i < len(word1):
6            result.append(word1[i])
7        if i < len(word2):
8            result.append(word2[i])
9        i += 1
10    return ''.join(result)

ℹ

Complexity note: This complexity is linear because we are iterating through both strings once, and the space complexity is also linear due to storing the result in a list before joining.

1Using two pointers allows for efficient traversal of both strings.
2Building the result with a list and joining it minimizes costly string concatenation operations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.