1754. Largest Merge Of Two Strings

Problem

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 (Java)

class Solution {
    public String largestMerge(String word1, String word2) {
        char[] a = word1.toCharArray();
        char[] b = word2.toCharArray();
        StringBuilder sb = new StringBuilder();
        int i = 0;
        int j = 0;
        while (i < a.length && j < b.length) {
            if (a[i] == b[j]) {
                boolean first = go(a, i, b, j);
                if (first) {
                    sb.append(a[i]);
                    i++;
                } else {
                    sb.append(b[j]);
                    j++;
                }
            } else {
                if (a[i] > b[j]) {
                    sb.append(a[i]);
                    i++;
                } else {
                    sb.append(b[j]);
                    j++;
                }
            }
        }

        while (i < a.length) {
            sb.append(a[i++]);
        }
        while (j < b.length) {
            sb.append(b[j++]);
        }

        return sb.toString();
    }

    private boolean go(char[] a, int i, char[] b, int j) {
        while (i < a.length && j < b.length && a[i] == b[j]) {
            i++;
            j++;
        }
        if (i == a.length) {
            return false;
        }
        if (j == b.length) {
            return true;
        }
        return a[i] > b[j];
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).