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 + ... + 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
func isInterleave(s1 string, s2 string, s3 string) bool {
if len(s3) != len(s1)+len(s2) {
return false
}
cache := make([][]*bool, len(s1)+1)
for i := range cache {
cache[i] = make([]*bool, len(s2)+1)
}
return isInterleaveHelper(s1, s2, s3, 0, 0, 0, cache)
}
func isInterleaveHelper(s1, s2, s3 string, i1, i2, i3 int, cache [][]*bool) bool {
if cache[i1][i2] != nil {
return *cache[i1][i2]
}
if i1 == len(s1) && i2 == len(s2) && i3 == len(s3) {
return true
}
result := false
if i1 < len(s1) && s1[i1] == s3[i3] {
result = isInterleaveHelper(s1, s2, s3, i1+1, i2, i3+1, cache)
}
if i2 < len(s2) && s2[i2] == s3[i3] {
result = result || isInterleaveHelper(s1, s2, s3, i1, i2+1, i3+1, cache)
}
cache[i1][i2] = &result
return result
}