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 they are divided into non-empty substrings such that:
s = s1 + s2 + ... + sn
t = t1 + t2 + ... + tm
|n - m| <= 1
s1 + t1 + s2 + t2 + s3 + t3 + ...
or t1 + 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
Example 2:
Input: s1 = “aabcc”, s2 = “dbbca”, s3 = “aadbbbaccc”
Output: false
Example 3:
Input: s1 = “”, s2 = “”, s3 = “”
Output: true
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?
public class Solution {
public bool IsInterleave(string s1, string s2, string s3) {
if (s1.Length + s2.Length != s3.Length) {
return false;
}
bool[,] dp = new bool[s1.Length + 1, s2.Length + 1];
dp[0, 0] = true;
for (int i = 0; i <= s1.Length; i++) {
for (int j = 0; j <= s2.Length; j++) {
if (i == 0 && j == 0) {
continue;
}
if (i > 0 && s1[i - 1] == s3[i + j - 1]) {
dp[i, j] |= dp[i - 1, j];
}
if (j > 0 && s2[j - 1] == s3[i + j - 1]) {
dp[i, j] |= dp[i, j - 1];
}
}
}
return dp[s1.Length, s2.Length];
}
}