Question

1754. Largest Merge Of Two Strings

中文文档

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

Solution 1

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)

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;
  }
};

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
  }
}

char* largestMerge(char* word1, char* word2) {
  int m = strlen(word1);
  int n = strlen(word2);
  int i = 0;
  int j = 0;
  char* ans = malloc((m + n + 1) * sizeof(char));
  while (i < m && j < n) {
    int k = 0;
    while (word1[i + k] && word2[j + k] && word1[i + k] == word2[j + k]) {
      k++;
    }
    if (word1[i + k] > word2[j + k]) {
      ans[i + j] = word1[i];
      i++;
    } else {
      ans[i + j] = word2[j];
      j++;
    };
  }
  while (word1[i]) {
    ans[i + j] = word1[i];
    i++;
  }
  while (word2[j]) {
    ans[i + j] = word2[j];
    j++;
  }
  ans[m + n] = '\0';
  return ans;
}