Skip to content

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

Solution 1

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
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)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
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();
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
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()
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
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;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
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
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
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;
}

Comments