What is the time complexity of Largest Merge Of Two Strings?

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

What is the space complexity of Largest Merge Of Two Strings?

The space complexity is O(m+n).

What data structures are used to solve Largest Merge Of Two Strings?

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

What are the key insights for solving Largest Merge Of Two Strings?

Insight 1: The core idea is a greedy approach. At each step, we choose the larger of the current characters from `word1` and `word2`. Insight 2: The crucial part is handling ties. When the current characters are equal, we need to compare the remaining suffixes of `word1` and `word2` to make the optimal choice.

Which companies ask LeetCode 1754. Largest Merge Of Two Strings in interviews?

This problem is frequently asked at Snap.

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Largest Merge Of Two Strings - LeetCode 1754 Solution

Largest Merge Of Two Strings - Complete Solution Guide

Largest Merge Of Two Strings is LeetCode problem 1754, a Medium 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 . You want to construct a string merge in the following way: while either word1 or word2 are non-empty, choose one of the following options: If word1 is non-empty, append the first character in word1 to merge and delete it from word1 . For example, if word1 = "abc" and merge = "dv" , then after choosing this operation, word1 = "bc" and merge = "dva" . If word2 is non-empty, append the first character in word2 to merge and delete it from word2 . For examp

Detailed Explanation

The problem asks us to merge two strings, `word1` and `word2`, into a single string called `merge`. We must create the lexicographically *largest* possible `merge`. At each step, we can either take the first character from `word1` or `word2` and append it to `merge`. The goal is to maximize the lexicographical order of the final merged string. The constraints specify that the strings consist of lowercase English letters and have lengths between 1 and 3000.

Solution Approach

The solution uses a greedy approach with two pointers, `i` and `j`, iterating through `word1` and `word2` respectively. At each step, it compares `word1[i]` and `word2[j]`. If one is larger, its character is appended to the result, and its pointer is incremented. If they are equal, the remaining suffixes of the strings starting at those indices are compared. The character from the string with the larger suffix is appended, and its pointer is incremented. Once one string is exhausted, the remaining part of the other string is appended to the result.

Step-by-Step Algorithm

Step 1: Initialize an empty string `res` to store the merged string.
Step 2: Initialize two pointers, `i` to 0 for `word1` and `j` to 0 for `word2`.
Step 3: Iterate while both `i` is within the bounds of `word1` and `j` is within the bounds of `word2`.
Step 4: Inside the loop, compare `word1[i]` and `word2[j]`.
Step 5: If `word1[i] > word2[j]`, append `word1[i]` to `res` and increment `i`.
Step 6: If `word2[j] > word1[i]`, append `word2[j]` to `res` and increment `j`.
Step 7: If `word1[i] == word2[j]`, compare the suffixes `word1[i:]` and `word2[j:]`. If `word1[i:] > word2[j:]`, append `word1[i]` to `res` and increment `i`. Otherwise, append `word2[j]` to `res` and increment `j`.
Step 8: After the loop, if there are remaining characters in `word1`, append the substring `word1[i:]` to `res`.
Step 9: If there are remaining characters in `word2`, append the substring `word2[j:]` to `res`.
Step 10: Return the `res` string.

Key Insights

Insight 1: The core idea is a greedy approach. At each step, we choose the larger of the current characters from `word1` and `word2`.
Insight 2: The crucial part is handling ties. When the current characters are equal, we need to compare the remaining suffixes of `word1` and `word2` to make the optimal choice.
Insight 3: The use of string comparison (`word1[i:] > word2[j:]` or equivalent in other languages) directly leverages the built-in lexicographical comparison functionality.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m+n)

Topics

This problem involves: Two Pointers, String, Greedy.

Companies

Asked at: Snap.

Largest Merge Of Two Strings | LeetCode 1754

Medium

Two Pointers String Greedy

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

Solution Code

Available in:

class Solution {
    public String largestMerge(String word1, String word2) {
        StringBuilder res = new StringBuilder();
        int i = 0, j = 0;
        int n1 = word1.length(), n2 = word2.length();

        while (i < n1 && j < n2) {
            if (word1.charAt(i) > word2.charAt(j)) {
                res.append(word1.charAt(i));
                i++;
            } else if (word2.charAt(j) > word1.charAt(i)) {
                res.append(word2.charAt(j));
                j++;
            } else {
                // Tie-breaking by comparing the remaining suffixes
                if (word1.substring(i).compareTo(word2.substring(j)) > 0) {
                    res.append(word1.charAt(i));
                    i++;
                } else {
                    res.append(word2.charAt(j));
                    j++;
                }
            }
        }

        if (i < n1) {
            res.append(word1.substring(i));
        }

        if (j < n2) {
            res.append(word2.substring(j));
        }

        return res.toString();
    }
}

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Problem Info

Description

Problem 1754: Largest Merge Of Two Strings - Detailed explanation and solution approaches.

Topics

Two Pointers String Greedy

Companies

Snap

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1754

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

10 steps

1
Step 1: Initialize an empty string `res` to store the merged string.
2
Step 2: Initialize two pointers, `i` to 0 for `word1` and `j` to 0 for `word2`.
3
Step 3: Iterate while both `i` is within the bounds of `word1` and `j` is within the bounds of `word2`.
4
Step 4: Inside the loop, compare `word1[i]` and `word2[j]`.
5
Step 5: If `word1[i] > word2[j]`, append `word1[i]` to `res` and increment `i`.
6
Step 6: If `word2[j] > word1[i]`, append `word2[j]` to `res` and increment `j`.
7
Step 7: If `word1[i] == word2[j]`, compare the suffixes `word1[i:]` and `word2[j:]`. If `word1[i:] > word2[j:]`, append `word1[i]` to `res` and increment `i`. Otherwise, append `word2[j]` to `res` and increment `j`.
8
Step 8: After the loop, if there are remaining characters in `word1`, append the substring `word1[i:]` to `res`.
9
Step 9: If there are remaining characters in `word2`, append the substring `word2[j:]` to `res`.
10
Step 10: Return the `res` string.

💡

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 time complexity is dominated by the suffix comparisons that occur when the characters at the current pointers are equal. The space complexity is determined by the length of the merged string, which will at most be the sum of the lengths of the input strings.

📖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

Dynamic Programming

A dynamic programming approach could be used to precompute the lexicographically largest suffix starting at each index of both strings. This would allow for O(1) lookup of the suffix comparison results, but it would add the overhead of building the DP table first.

Time: O(m*n) for building the DP table and O(m+n) to construct the merge, so O(m*n) overall.Space: O(m*n) for the DP table + O(m+n) for the result, so O(m*n) overall.

Trade-offs: This approach is less efficient in terms of both time and space compared to the greedy approach. The greedy approach is much more straightforward to implement and achieves the same results in the same order of complexity.

✨

Key Insights

1Insight 1: The core idea is a greedy approach. At each step, we choose the larger of the current characters from `word1` and `word2`.
2Insight 2: The crucial part is handling ties. When the current characters are equal, we need to compare the remaining suffixes of `word1` and `word2` to make the optimal choice.
3Insight 3: The use of string comparison (`word1[i:] > word2[j:]` or equivalent in other languages) directly leverages the built-in lexicographical comparison functionality.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

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

Edge Cases

Edge case 1: One string is empty. The algorithm handles this correctly because the while loop condition `i < n1 and j < n2` will terminate when one of the strings is exhausted, and the remaining characters from the other string will be appended.
Edge case 2: Both strings are the same. The algorithm correctly handles this by comparing suffixes. For example, if both are "abc", it will compare "abc" with "abc", select from either and continue.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case when the current characters from both strings are equal. Simply choosing one string arbitrarily will not guarantee the lexicographically largest merge.

✗Common mistake 2: Forgetting to append the remaining characters from the non-empty string after one string is exhausted. The while loop only considers the case where both strings have characters remaining.

✅

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?

💼Real-world scenario 1: String compression algorithms might use similar logic for identifying and merging repeated or related substrings to reduce the overall string length.

🏦Real-world scenario 2: In bioinformatics, sequence alignment algorithms often compare and merge DNA or protein sequences, using similar techniques to find the optimal alignment based on some scoring function (analogous to lexicographical order).

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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 13, 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

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

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Largest Merge Of Two Strings"?

For this medium-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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