tomatocream
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Add populate-notes.mjs that fetches problem descriptions and Python/C++ code stubs from LeetCode's GraphQL API. Populated all 197 NeetCode 150 note files with: - Problem description (examples, constraints) - Python code stub (function signature) - C++ code stub (function signature + includes) API responses cached in leetcode/.cache/leetcode/ for instant re-runs.
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1.9 KiB
TODO 0097. 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 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} + ...ort_{1} + s_{1} + t_{2} + s_{2} + t_{3} + s_{3} + ...
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: trueConstraints:
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?
TODO Approach
Write your approach here.
TODO Python
class Solution:
def isInterleave(self, s1: str, s2: str, s3: str) -> bool:TODO C++
class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
}
};