Largest Merge Of Two Strings

Largest Merge Of Two Strings - Problem

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.
If word2 is non-empty, append the first character in word2 to merge and delete it from word2.

Return the lexicographically largest merge you can construct.

A string a is lexicographically larger than a string b (of the same length) if in the first position where a and b differ, a has a character strictly larger than the corresponding character in b.

Input & Output

Example 1 — Basic Case

$ Input: word1 = "cab", word2 = "ab"

› Output: "cabab"

💡 Note: At each step: "cab" vs "ab" → choose 'c', then "ab" vs "ab" → choose 'a', then "b" vs "b" → choose 'a', append remaining "bb"

Example 2 — Equal Start

$ Input: word1 = "abcabc", word2 = "abdaba"

› Output: "abdabacabcabc"

💡 Note: "abcabc" vs "abdaba" → "abdaba" is larger, so choose 'a' from word2. Continue comparing suffixes.

Example 3 — One Empty

$ Input: word1 = "a", word2 = "b"

› Output: "ba"

💡 Note: "a" vs "b" → 'b' > 'a', so choose 'b' first, then append remaining 'a'

Constraints

1 ≤ word1.length, word2.length ≤ 3000
word1 and word2 consist only of lowercase English letters

Visualization

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Asked in

G Google 25 a Amazon 18 M Microsoft 12

The key insight is to use a greedy approach: at each step, choose the character from the string whose remaining suffix is lexicographically larger. This ensures we always make the optimal choice for the largest possible merge. Best approach is greedy suffix comparison. Time: O((m+n)²), Space: O(1)

Common Approaches

✓ Dfs

⏱️ Time: N/A Space: N/A

Brute Force Recursion

⏱️ Time: O(2^(m+n)) Space: O(m+n)

Use recursion to explore every possible choice at each step. At each position, we can take from either word1 or word2, so we try both options and return the lexicographically larger result.

Greedy Choice

⏱️ Time: O((m+n)²) Space: O(1)

Make greedy decisions by always choosing from the string that has lexicographically larger remaining suffix. When current characters are equal, look ahead to make the optimal choice.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Smart Actions

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