c++ permutation
The 3! possible permutations with 3 elements:
1 2 4
31 69
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
After loop: 1 2 3c++ permutation
The 3! possible permutations with 3 elements:
1 2 4
31 69
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
After loop: 1 2 3permutation in a string
import collections
class Solution:
def checkInclusion(self, s1, s2):
"""
:type s1: str
:type s2: str
:rtype: bool
"""
'''
Apprach 3:
Just changing the collections.Counter to collections.defaultdict
improved the speed
94.36%, Nice. :)
'''
ctr1 = collections.defaultdict(int)
ctr2 = collections.defaultdict(int)
for x in s1: ctr1[x] += 1
for x in s2[:len(s1)]: ctr2[x] += 1
i = 0; j = len(s1)
while j < len(s2):
if ctr2 == ctr1: return True
ctr2[s2[i]] -= 1
if ctr2[s2[i]] < 1: ctr2.pop(s2[i]);
ctr2[s2[j]] = ctr2.get(s2[j], 0) + 1
i += 1; j += 1
return ctr2 == ctr1
'''
Approach 2:
Use the same counter but much more effeciently
63% beated, okay not bad..
two pointers i(front) j(last), delete and add accordingly and check for the counts
# Code Below
'''
ctr1 = collections.Counter(s1)
ctr2 = collections.Counter(s2[:len(s1)])
i = 0; j = len(s1)
while j < len(s2):
if ctr2 == ctr1: return True
ctr2[s2[i]] -= 1
if ctr2[s2[i]] < 1: ctr2.pop(s2[i]);
ctr2[s2[j]] = ctr2.get(s2[j], 0) + 1
i += 1; j += 1
return ctr2 == ctr1
'''
1. Approach 1
Things i Missed:
- Had the len(s2) - len(s1) + 1: forgot to add the 1 in the if condition
Ultra easy solution
Slice and check for the equality
5% beat, yuck !!
# Code below
'''
ctr1 = collections.Counter(s1)
i = 0
while i < len(s2) - len(s1) + 1:
if s2[i] in ctr1:
ctr2 = collections.Counter(s2[i: i+len(s1)])
if ctr1 == ctr2: return True
i += 1
return FalseCopyright © 2021 Codeinu
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