Class Solution
-
- All Implemented Interfaces:
public final class Solution
97 - Interleaving String.
Medium
Given strings
s1
,s2
, ands3
, find whethers3
is formed by an interleaving ofs1
ands2
.An interleaving of two strings
s
andt
is a configuration wheres
andt
are divided inton
andm
non-empty substrings respectively, such that:<code>s = s<sub>1</sub> + s<sub>2</sub> + ... + s<sub>n</sub></code>
<code>t = t<sub>1</sub> + t<sub>2</sub> + ... + t<sub>m</sub></code>
|n - m| <= 1
The interleaving is <code>s<sub>1</sub> + t<sub>1</sub> + s<sub>2</sub> + t<sub>2</sub> + s<sub>3</sub> + t<sub>3</sub> + ...</code> or <code>t<sub>1</sub> + s<sub>1</sub> + t<sub>2</sub> + s<sub>2</sub> + t<sub>3</sub> + s<sub>3</sub> + ...</code>
Note:
a + b
is the concatenation of stringsa
andb
.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
, ands3
consist of lowercase English letters.
Follow up: Could you solve it using only
O(s2.length)
additional memory space?
-
-
Constructor Summary
Constructors Constructor Description Solution()
-