g1701_1800.s1754_largest_merge_of_two_strings.Solution Maven / Gradle / Ivy

Go to download
Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation
Java-based LeetCode algorithm problem solutions, regularly updated
There is a newer version: 1.38
package g1701_1800.s1754_largest_merge_of_two_strings;

// #Medium #String #Greedy #Two_Pointers #2022_04_30_Time_37_ms_(89.23%)_Space_53.8_MB_(53.08%)

/**
 * 1754 - Largest Merge Of Two Strings\.
 *
 * Medium
 *
 * 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.
**/
public 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];
    }
}