LeetCode Top Interview 150

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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