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 + ... + snt = t1 + t2 + ... + tm|n - m| <= 1- The interleaving is
s1 + t1 + s2 + t2 + s3 + t3 + ...ort1 + 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 <= 1000 <= s3.length <= 200s1,s2, ands3consist 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
}
}