---
description: Greedy
---

# LeetCode 1754. Largest Merge Of Two Strings



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.

## Solution:

[English Version in Youtube](https://youtu.be/l5gKIfFbCIY)

[中文版解答Youtube Link](https://youtu.be/69YytzC_g_w)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1C541177NK/)

A short version

```
class Solution {
    
public:
    string largestMerge(string word1, string word2) {
        string ans;
        int i = 0, j = 0;
        while(i < word1.size() && j < word2.size()) {
            if(word1.substr(i) > word2.substr(j)) {
                ans.push_back(word1[i]);
                i++;
            } else {
                ans.push_back(word2[j]);
                j++;
            }
        }
        
        if(i == word1.size()) {
            ans += word2.substr(j);
        } else if(j == word2.size()) {
            ans += word1.substr(i);
        }
        
        return ans;
    }
};
```

A faster version

```
class Solution {
    
    bool greater(const string &word1, const string &word2, int i, int j) {
        while (i < word1.size() && j < word2.size()) {
            if (word1[i] > word2[j]) return true;
            if (word1[i] < word2[j]) return false;
            i++;
            j++;
        }
        return (word1.size()-i) >= (word2.size()-j);
    }
    
public:
    string largestMerge(string word1, string word2) {
        string ans;
        int i = 0, j = 0;
        while(i < word1.size() && j < word2.size()) {
            if(greater(word1, word2, i, j)) {
                ans.push_back(word1[i]);
                i++;
            } else {
                ans.push_back(word2[j]);
                j++;
            }
        }
        if(i == word1.size()) {
            ans += word2.substr(j);
        } else if(j == word2.size()) {
            ans += word1.substr(i);
        }
        
        return ans;
    }
};
```
