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 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:

English Version in Youtube

中文版解答Youtube Link

中文版解答Bilibili Link

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;
    }
};