LeetCode 1754. Largest Merge Of Two Strings
Greedy
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 inword1
tomerge
and delete it fromword1
.For example, if
word1 = "abc"
andmerge = "dv"
, then after choosing this operation,word1 = "bc"
andmerge = "dva"
.
If
word2
is non-empty, append the first character inword2
tomerge
and delete it fromword2
.For example, if
word2 = "abc"
andmerge = ""
, then after choosing this operation,word2 = "bc"
andmerge = "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
andword2
consist only of lowercase English letters.
Solution:
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;
}
};
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