"""
97. Interleaving String
Description
HintsSubmissionsDiscussSolution
Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
For example,
Given:
s1 = "aabcc",
s2 = "dbbca",
When s3 = "aadbbcbcac", return true.
When s3 = "aadbbbaccc", return false.
"""
class Solution:
# BFS
def isInterleave(self, s1, s2, s3):
lx, ly, lz = len(s1), len(s2), len(s3)
if lx + ly != lz:
return False
haved = [(0, 0)]
visitedx = set()
visitedy = set()
while haved:
x, y = haved.pop(0)
if x + y == lz:
return True
# print(x,y)
# if x<lx and s1[x]==s3[x+y] and (x,y) not in visited :
if x < lx and s1[x] == s3[x + y] and (x, y) not in visitedx:
haved.append((x + 1, y))
visitedx.add((x, y))
# print('x',x,y)
# if y<ly and s2[y]==s3[x+y] and (x,y) not in visited:
if y < ly and s2[y] == s3[x + y] and (x, y) not in visitedy:
haved.append((x, y + 1))
visitedy.add((x, y))
# print('y',x,y)
return False
if __name__ == '__main__':
s1 = "a"
s2 = ""
s3 = "a"
# s1 = "aabcc"
# s2 = "dbbcc"
# aabc
# dbbc
# s3 = "aadbbcbcac"
# s3 = "aadbbbaccc",
s1 = "a"
s2 = "b"
s3 = "a"
s1 = "abbbbbbcabbacaacccababaabcccabcacbcaabbbacccaaaaaababbbacbb"
s2 = "ccaacabbacaccacababbbbabbcacccacccccaabaababacbbacabbbbabc"
s3 = "cacbabbacbbbabcbaacbbaccacaacaacccabababbbababcccbabcabbaccabcccacccaabbcbcaccccaaaaabaaaaababbbbacbbabacbbacabbbbabc"
s1 = "aa"
s2 = "ab"
s3 = "aaba"
st = Solution()
out = st.isInterleave(s1, s2, s3)
print(out)
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