LeetCode #1754 — MEDIUM

Largest Merge Of Two Strings

Move from brute-force thinking to an efficient approach using two pointers strategy.

The Problem

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 example, if word2 = "abc" and merge = "", then after choosing this operation, word2 = "bc" and merge = "a".

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. For example, "abcd" is lexicographically larger than "abcc" because the first position they differ is at the fourth character, and d is greater than c.

Example 1:

Input: word1 = "cabaa", word2 = "bcaaa"
Output: "cbcabaaaaa"
Explanation: One way to get the lexicographically largest merge is:
- Take from word1: merge = "c", word1 = "abaa", word2 = "bcaaa"
- Take from word2: merge = "cb", word1 = "abaa", word2 = "caaa"
- Take from word2: merge = "cbc", word1 = "abaa", word2 = "aaa"
- Take from word1: merge = "cbca", word1 = "baa", word2 = "aaa"
- Take from word1: merge = "cbcab", word1 = "aa", word2 = "aaa"
- Append the remaining 5 a's from word1 and word2 at the end of merge.

Example 2:

Input: word1 = "abcabc", word2 = "abdcaba"
Output: "abdcabcabcaba"

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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 example, if word2 = "abc" and merge = "", then after choosing this operation, word2 = "bc" and merge = "a". 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. For example, "abcd" is lexicographically

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Two Pointers · Greedy

Example 1

"cabaa"
"bcaaa"

Example 2

"abcabc"
"abdcaba"

Core Insight

What unlocks the optimal approach

Build the result character by character. At each step, you choose a character from one of the two strings.
If the next character of the first string is larger than that of the second string, or vice versa, it's optimal to use the larger one.
If both are equal, think of a criteria that lets you decide which string to consume the next character from.
You should choose the next character from the larger string.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1754: Largest Merge Of Two Strings
class Solution {
    public String largestMerge(String word1, String word2) {
        int m = word1.length(), n = word2.length();
        int i = 0, j = 0;
        StringBuilder ans = new StringBuilder();
        while (i < m && j < n) {
            boolean gt = word1.substring(i).compareTo(word2.substring(j)) > 0;
            ans.append(gt ? word1.charAt(i++) : word2.charAt(j++));
        }
        ans.append(word1.substring(i));
        ans.append(word2.substring(j));
        return ans.toString();
    }
}

// Accepted solution for LeetCode #1754: Largest Merge Of Two Strings
func largestMerge(word1 string, word2 string) string {
	m, n := len(word1), len(word2)
	i, j := 0, 0
	var ans strings.Builder
	for i < m && j < n {
		if word1[i:] > word2[j:] {
			ans.WriteByte(word1[i])
			i++
		} else {
			ans.WriteByte(word2[j])
			j++
		}
	}
	ans.WriteString(word1[i:])
	ans.WriteString(word2[j:])
	return ans.String()
}

# Accepted solution for LeetCode #1754: Largest Merge Of Two Strings
class Solution:
    def largestMerge(self, word1: str, word2: str) -> str:
        i = j = 0
        ans = []
        while i < len(word1) and j < len(word2):
            if word1[i:] > word2[j:]:
                ans.append(word1[i])
                i += 1
            else:
                ans.append(word2[j])
                j += 1
        ans.append(word1[i:])
        ans.append(word2[j:])
        return "".join(ans)

// Accepted solution for LeetCode #1754: Largest Merge Of Two Strings
impl Solution {
    pub fn largest_merge(word1: String, word2: String) -> String {
        let word1 = word1.as_bytes();
        let word2 = word2.as_bytes();
        let m = word1.len();
        let n = word2.len();
        let mut ans = String::new();
        let mut i = 0;
        let mut j = 0;
        while i < m && j < n {
            if word1[i..] > word2[j..] {
                ans.push(word1[i] as char);
                i += 1;
            } else {
                ans.push(word2[j] as char);
                j += 1;
            }
        }
        word1[i..].iter().for_each(|c| ans.push(*c as char));
        word2[j..].iter().for_each(|c| ans.push(*c as char));
        ans
    }
}

// Accepted solution for LeetCode #1754: Largest Merge Of Two Strings
function largestMerge(word1: string, word2: string): string {
    const m = word1.length;
    const n = word2.length;
    let ans = '';
    let i = 0;
    let j = 0;
    while (i < m && j < n) {
        ans += word1.slice(i) > word2.slice(j) ? word1[i++] : word2[j++];
    }
    ans += word1.slice(i);
    ans += word2.slice(j);
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.