Problem
Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
Example 1:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
Example 2:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false
Solution
DP problem.
Imagin you have a matrix with s1 as its row and s2 as its column.
You have to find a path from upper left corner to bottom right corner, which consisits the s3.
So when you walk the path, you have two options:
- you come from left. dp[i][j] = dp[i][j-1] and s1[j-1] == s3[i+j-1]
- you come from above. dp[i][j] = dp[i][j-1] and s2[i-1] == s3[i+j-1]
class Solution:
def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
if len(s1) + len(s2) != len(s3):
return False
dp = [[False] * (len(s1)+1) for _ in range(len(s2)+1)]
for i in range(len(s2)+1):
for j in range(len(s1)+1):
if not i and not j:
dp[i][j] = True
elif not i:
dp[i][j] = dp[0][j-1] and s1[j-1] == s3[j-1]
elif not j:
dp[i][j] = dp[i-1][0] and s2[i-1] == s3[i-1]
else:
dp[i][j] = (dp[i-1][j] and s2[i-1] == s3[i+j-1]) or (dp[i][j-1] and s1[j-1] == s3[i+j-1])
return dp[-1][-1]