LeetCode in Kotlin

97. Interleaving String

Medium

Given strings s1, s2, and s3, find whether s3 is formed by an interleaving of s1 and s2.

An interleaving of two strings s and t is a configuration where s and t are divided into n and m non-empty substrings respectively, such that:

• s = s1 + s2 + ... + sn
• t = t1 + t2 + ... + tm
• |n - m| <= 1
• The interleaving is s1 + t1 + s2 + t2 + s3 + t3 + ... or t1 + s1 + t2 + s2 + t3 + s3 + ...

Note: a + b is the concatenation of strings a and b.

Example 1:

Input: s1 = “aabcc”, s2 = “dbbca”, s3 = “aadbbcbcac”

Output: true

Explanation: One way to obtain s3 is: Split s1 into s1 = “aa” + “bc” + “c”, and s2 into s2 = “dbbc” + “a”. Interleaving the two splits, we get “aa” + “dbbc” + “bc” + “a” + “c” = “aadbbcbcac”. Since s3 can be obtained by interleaving s1 and s2, we return true.

Example 2:

Input: s1 = “aabcc”, s2 = “dbbca”, s3 = “aadbbbaccc”

Output: false

Explanation: Notice how it is impossible to interleave s2 with any other string to obtain s3.

Example 3:

Input: s1 = “”, s2 = “”, s3 = “”

Output: true

Constraints:

• 0 <= s1.length, s2.length <= 100
• 0 <= s3.length <= 200
• s1, s2, and s3 consist of lowercase English letters.

Follow up: Could you solve it using only O(s2.length) additional memory space?

Solution

class Solution {
    fun isInterleave(s1: String, s2: String, s3: String): Boolean {
        if (s3.length != s1.length + s2.length) {
            return false
        }
        val cache = Array(s1.length + 1) { arrayOfNulls<Boolean>(s2.length + 1) }
        return isInterleave(s1, s2, s3, 0, 0, 0, cache)
    }

    fun isInterleave(
        s1: String,
        s2: String,
        s3: String,
        i1: Int,
        i2: Int,
        i3: Int,
        cache: Array<Array<Boolean?>>,
    ): Boolean {
        if (cache[i1][i2] != null) {
            return cache[i1][i2]!!
        }
        if (i1 == s1.length && i2 == s2.length && i3 == s3.length) {
            return true
        }
        var result = false
        if (i1 < s1.length && s1[i1] == s3[i3]) {
            result = isInterleave(s1, s2, s3, i1 + 1, i2, i3 + 1, cache)
        }
        if (i2 < s2.length && s2[i2] == s3[i3]) {
            result = result || isInterleave(s1, s2, s3, i1, i2 + 1, i3 + 1, cache)
        }
        cache[i1][i2] = result
        return result
    }
}

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