LeetCode in Kotlin

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.

Solution

@Suppress("NAME_SHADOWING")
class Solution {
    fun largestMerge(word1: String, word2: String): String {
        val a = word1.toCharArray()
        val b = word2.toCharArray()
        val sb = StringBuilder()
        var i = 0
        var j = 0
        while (i < a.size && j < b.size) {
            if (a[i] == b[j]) {
                val 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.size) {
            sb.append(a[i++])
        }
        while (j < b.size) {
            sb.append(b[j++])
        }
        return sb.toString()
    }

    private fun go(a: CharArray, i: Int, b: CharArray, j: Int): Boolean {
        var i = i
        var j = j
        while (i < a.size && j < b.size && a[i] == b[j]) {
            i++
            j++
        }
        if (i == a.size) {
            return false
        }
        return if (j == b.size) {
            true
        } else {
            a[i] > b[j]
        }
    }
}

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