cpp-flashcards/org/study_deck_02/dsa/2-d-dynamic-programming/0097-interleaving-string.org

#+ANKI_DECK: study_deck_02
* TODO 0097. Interleaving String :medium:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*0097. Interleaving String][0097. Interleaving String]]
:END:

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~ substrings respectively, such that:

 - ~s = s_{1} + s_{2} + ... + s_{n}~

 - ~t = t_{1} + t_{2} + ... + t_{m}~

 - ~|n - m| <= 1~

 - The *interleaving* is ~s_{1} + t_{1} + s_{2} + t_{2} + s_{3} + t_{3} + ...~ or ~t_{1} + s_{1} + t_{2} + s_{2} + t_{3} + s_{3} + ...~

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

*Example 1:*


#+begin_src
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.
#+end_src


*Example 2:*


#+begin_src
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.
#+end_src


*Example 3:*


#+begin_src
Input: s1 = "", s2 = "", s3 = ""
Output: true
#+end_src


*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?

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python
class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
#+end_src

** TODO C++
#+begin_src cpp
class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        
    }
};
#+end_src