# 1754. Largest Merge Of Two Strings

## Description

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.

## Solutions

• 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();
}
}

• class Solution {
public:
string largestMerge(string word1, string word2) {
int m = word1.size(), n = word2.size();
int i = 0, j = 0;
string ans;
while (i < m && j < n) {
bool gt = word1.substr(i) > word2.substr(j);
ans += gt ? word1[i++] : word2[j++];
}
ans += word1.substr(i);
ans += word2.substr(j);
return ans;
}
};

• 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)


• 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()
}

• 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;
}


• 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
}
}