# DP in python with explaination

• c1,c2,c3 as the pointer to s1,s2,s3
if c1 is ok and c2 is ok, choose c1 first,keep going
if stucked, suppose j in s3, find all s3[j] in (c2,min(len(s2),c2+c1+j) save in list p
for every position in p, dp check
s1[:c1-c2+j] , s2[:j], s3[:c3] are Ture
and s1[c1-c2+j],s2[j],s3[c3] are True

``````class Solution(object):
def isInterleave(self, s1, s2, s3):
"""
:type s1: str
:type s2: str
:type s3: str
:rtype: bool
"""
c1,c2,c3=0,0,0
if s1=="":return s2==s3
elif s2=="":return s1==s3
elif s3=="":return False

if len(s3)!=len(s1)+len(s2):return False
for c3 in range(0,len(s3)):
i=s3[c3]
if c1<len(s1) and i==s1[c1] :c1+=1
elif c2<len(s2) and i==s2[c2] :c2+=1
else:
#j=-1
jj=[]
for x in range(c2+1,min(len(s2),c1+c2+1)):
if s2[x]==i:
jj.append(x)
if len(jj)==0:return False
else:
for j in jj:
if self.isInterleave(s1[:c1-j+c2],s2[:j],s3[:c3]):
c1=c1-j+c2
c2=j
if self.isInterleave(s1[c1:],s2[j+1:],s3[c3+1:]):return True
return False
#else:return False
return True
``````

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