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Formatted question description: https://leetcode.ca/all/1754.html
1754. Largest Merge Of Two Strings
Level
Medium
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 inword1
tomerge
and delete it fromword1
.- For example, if
word1 = "abc"
andmerge = "dv"
, then after choosing this operation,word1 = "bc"
andmerge = "dva"
.
- For example, if
- 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"
.
- For example, if
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
Use two pointers index1
and index2
to point to indices of word1
and word2
, respectively. Each time compare the two characters at index1
and index2
. If the two characters are different, append the larger character to merge
. Otherwise, compare the remaining substrings of word1
and word2
starting from index1
and index2
respectively, and select the first character of the greater substring and append it to merge
. Finally, return merge
.
-
class Solution { public String largestMerge(String word1, String word2) { StringBuffer merge = new StringBuffer(); int length1 = word1.length(), length2 = word2.length(); int index1 = 0, index2 = 0; while (index1 < length1 && index2 < length2) { char c1 = word1.charAt(index1), c2 = word2.charAt(index2); if (c1 > c2) { merge.append(c1); index1++; } else if (c1 < c2) { merge.append(c2); index2++; } else { int temp1 = index1 + 1, temp2 = index2 + 1; boolean flag = true; while (temp1 < length1 && temp2 < length2) { if (word1.charAt(temp1) == word2.charAt(temp2)) { temp1++; temp2++; } else if (word1.charAt(temp1) > word2.charAt(temp2)) { merge.append(c1); index1++; flag = false; break; } else { merge.append(c2); index2++; flag = false; break; } } if (flag) { if (temp2 == length2) { merge.append(c1); index1++; } else { merge.append(c2); index2++; } } } } while (index1 < length1) { merge.append(word1.charAt(index1)); index1++; } while (index2 < length2) { merge.append(word2.charAt(index2)); index2++; } return merge.toString(); } } ############ 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(); } }
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// OJ: https://leetcode.com/problems/largest-merge-of-two-strings/ // Time: O(N^2) // Space: O(N) class Solution { public: string largestMerge(string word1, string word2) { priority_queue<string> q; q.push(word1); q.push(word2); string ans; while (q.size()) { auto s = q.top(); q.pop(); ans += s[0]; if (s.size() > 1) q.push(s.substr(1)); } return ans; } };
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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) ############ # 1754. Largest Merge Of Two Strings # https://leetcode.com/problems/largest-merge-of-two-strings class Solution: def largestMerge(self, word1: str, word2: str) -> str: res = "" while word1 and word2: if word1 >= word2: res += word1[0] word1 = word1[1:] else: res += word2[0] word2 = word2[1:] res += word1 or word2 return res
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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() }
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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; }
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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 } }