Largest Merge Of Two Strings

Largest Merge Of Two Strings

1. What is this problem about?

The Largest Merge Of Two Strings coding problem presents you with two strings, word1 and word2. Your goal is to construct a third string, merge, by repeatedly picking the first character from either word1 or word2 and appending it to merge. To make the merge the "lexicographically largest," you must strategically decide which character to pick at each step so that the resulting string appears as late as possible in an alphabetical dictionary.

2. Why is this asked in interviews?

This question is frequently asked at companies like Snap and Snapchat to test a candidate's understanding of greedy algorithms and string comparison mechanics. It requires more than just comparing the first characters; you often have to look ahead when the leading characters are identical. It evaluates your ability to handle "tie-breaking" scenarios efficiently, which is a common requirement in real-world data processing and sorting tasks.

3. Algorithmic pattern used

This problem follows the Two Pointers and Greedy interview pattern. At each step, you compare the remaining suffixes of both strings. The greedy choice is to take the character from the string whose current suffix is lexicographically larger. This ensure that you are always making the best local decision to achieve the best global outcome. String slicing or pointer manipulation is used to perform these comparisons.

4. Example explanation

Suppose word1 = "banana" and word2 = "bandana". Step 1: Compare "banana" and "bandana". "bandana" is larger (at the 4th char 'd' > 'a'). Take 'b' from "bandana". Step 2: Compare "banana" and "andana". "banana" is larger. Take 'b' from "banana". Step 3: Compare "anana" and "andana". "andana" is larger. Take 'a' from "andana". This greedy selection continues until one string is empty, at which point you append the remainder of the other string.

5. Common mistakes candidates make

The most common pitfall is only comparing the immediate first characters (word1[i] vs word2[j]). If they are equal, simply picking from either string at random will fail. You must compare the entire remaining substrings to decide. Another mistake is using inefficient string concatenation (like s = s + char) in languages where strings are immutable, which can lead to O(N²) time complexity.

6. Interview preparation tip

For "String, Greedy interview pattern" problems, always consider the edge cases where prefixes are identical. Practice using built-in string comparison functions, as they are often highly optimized. Remember that lexicographical order is just like dictionary order: "apple" comes before "apply" because 'e' comes before 'y'.