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# Largest Merge Of Two Strings

Updated: Mar 24, 2021

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:

```class Solution {
public String largestMerge(String word1, String word2) {
if(word1.length()==0 && word2.length()==0)
return "";
if(word1.length()==0) return word2;
if(word2.length()==0) return word1;

if(word2.compareTo(word1)<0)
{
return word1.substring(0,1)+largestMerge(word1.substring(1),word2);
}
else
{
return word2.substring(0,1)+largestMerge(word1,word2.substring(1));
}
}
}```