What is the time complexity of Merge Strings Alternately?

The optimal solution has a time complexity of O(m+n).

What is the space complexity of Merge Strings Alternately?

The space complexity is O(m+n).

What data structures are used to solve Merge Strings Alternately?

This problem uses Two Pointers, String. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Merge Strings Alternately?

The problem can be solved efficiently using two pointers, one for each string, to track the current character being processed. Using a `StringBuilder` (in Java) or similar techniques (like string concatenation in Python or C++) is crucial for optimal performance, especially for large strings. Repeated string concatenation can lead to significant performance overhead.

Which companies ask LeetCode 1768. Merge Strings Alternately in interviews?

This problem is frequently asked at Wells Fargo, Zoho.

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Merge Strings Alternately - LeetCode 1768 Solution

Merge Strings Alternately - Complete Solution Guide

Merge Strings Alternately is LeetCode problem 1768, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem asks you to merge two strings, `word1` and `word2`, alternately. Start by taking the first character from `word1`, then the first character from `word2`, and so on. If one string is longer than the other, append the remaining characters of the longer string to the end of the merged string. The output is the resulting merged string.

Solution Approach

The solution uses a two-pointer approach. Two pointers (`i` and `j`) iterate through `word1` and `word2` respectively. The algorithm repeatedly appends the character at the current pointer position from each string to the `merged` string, incrementing the pointers. After one string is exhausted, the algorithm appends the remaining characters of the other string to `merged`.

Step-by-Step Algorithm

Initialize two pointers, `i` and `j`, to 0. These pointers will track the indices in `word1` and `word2` respectively.
Initialize an empty string or StringBuilder called `merged` to store the result.
While `i` is less than the length of `word1` AND `j` is less than the length of `word2`:
Append `word1[i]` to `merged`.
Append `word2[j]` to `merged`.
Increment `i` and `j`.
While `i` is less than the length of `word1`:
Append `word1[i]` to `merged`.
Increment `i`.
While `j` is less than the length of `word2`:
Append `word2[j]` to `merged`.
Increment `j`.
Return `merged`.

Key Insights

The problem can be solved efficiently using two pointers, one for each string, to track the current character being processed.
Using a `StringBuilder` (in Java) or similar techniques (like string concatenation in Python or C++) is crucial for optimal performance, especially for large strings. Repeated string concatenation can lead to significant performance overhead.
Handling the cases where one string is longer than the other requires careful attention to ensure all characters from both strings are included in the merged string.

Complexity Analysis

Time Complexity: O(m+n)

Space Complexity: O(m+n)

Topics

This problem involves: Two Pointers, String.

Companies

Asked at: Wells Fargo, Zoho.

Merge Strings Alternately | LeetCode 1768

Easy

Two Pointers String

Time: O(m+n)Space: O(m+n)

Solution Code

Available in:

class Solution {
    public String mergeAlternately(String word1, String word2) {
        StringBuilder merged = new StringBuilder();
        int i = 0, j = 0;
        while (i < word1.length() && j < word2.length()) {
            merged.append(word1.charAt(i++));
            merged.append(word2.charAt(j++));
        }
        while (i < word1.length()) {
            merged.append(word1.charAt(i++));
        }
        while (j < word2.length()) {
            merged.append(word2.charAt(j++));
        }
        return merged.toString();
    }
}

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Problem Info

Description

Problem 1768: Merge Strings Alternately - Detailed explanation and solution approaches.

Topics

Two Pointers String

Companies

Wells FargoZoho

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:1768

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

13 steps

1
Initialize two pointers, `i` and `j`, to 0. These pointers will track the indices in `word1` and `word2` respectively.
2
Initialize an empty string or StringBuilder called `merged` to store the result.
3
While `i` is less than the length of `word1` AND `j` is less than the length of `word2`:
4
Append `word1[i]` to `merged`.
5
Append `word2[j]` to `merged`.
6
Increment `i` and `j`.
7
While `i` is less than the length of `word1`:
8
Append `word1[i]` to `merged`.
9
Increment `i`.
10
While `j` is less than the length of `word2`:
11
Append `word2[j]` to `merged`.
12
Increment `j`.
13
Return `merged`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m+n)

💾

Space Complexity

O(m+n)

📝

Why?

The algorithm iterates through each character of both strings once. In the worst case, it processes all m characters of `word1` and all n characters of `word2`. The space used by `merged` is proportional to the sum of the lengths of the input strings (m + n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Approach

A recursive solution could be implemented, although it's less efficient than the iterative approach. It would involve recursively merging prefixes of the strings and concatenating the results.

Time: O(m + n)Space: O(m + n) - due to recursive call stack

Trade-offs: The iterative approach is generally preferred due to its better space complexity and easier readability. Recursion can lead to stack overflow errors for extremely large input strings.

✨

Key Insights

1The problem can be solved efficiently using two pointers, one for each string, to track the current character being processed.
2Using a `StringBuilder` (in Java) or similar techniques (like string concatenation in Python or C++) is crucial for optimal performance, especially for large strings. Repeated string concatenation can lead to significant performance overhead.
3Handling the cases where one string is longer than the other requires careful attention to ensure all characters from both strings are included in the merged string.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Empty strings: If either `word1` or `word2` (or both) are empty, the algorithm correctly handles this by appending the remaining characters of the non-empty string.
One string significantly longer than the other: The algorithm correctly handles cases where one string is much longer than the other by appending the remaining characters of the longer string to the end.

⚠️

Common Mistakes to Avoid

✗Incorrectly handling the case where one string is longer than the other: Failing to append the remaining characters of the longer string after one string is fully processed.

✗Inefficient string concatenation: Repeatedly using the `+` operator in Python or similar string concatenation operations in other languages without using a `StringBuilder` or similar can lead to significant performance degradation for large strings.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Data merging: Merging two streams of data alternately, such as sensor readings from two different devices.

🏦Text processing: Combining strings from different sources in a specific pattern for output formatting or generating reports.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 27, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Merge Strings Alternately"?

For this easy-level Two Pointers problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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