Deepak Saini - 1 year ago 154

Python Question

I have been trying to implement tabular method for simplification of boolean expressions in python. For that I need to check whether two given strings differ at only one index

for example, the function should return the following for the following examples:

- and
`0011`

- true as the two differ only at position 1`0111`

- and
`0-001`

- true as differ at only 2`0-101`

- and
`0-011`

- false as differ at 2,3`0-101`

right now I am using the following function:

`def match(s1,s2):`

l=[False,-1]##returns false when they cant be combined

for i in range(len(s1)):

if s1[:i]==s2[:i] and s1[i]!=s2[i] and s1[i+1:]==s2[i+1:]:

l= [True,i]

break

return l

I want to implement it in a very fast manner (low complexity). Is there a way to do so in python?

Recommended for you: Get network issues from **WhatsUp Gold**. **Not end users.**

Answer Source

This is a more well-performing solution, coded in Python 3:

```
def match(s1, s2):
ok = False
for c1, c2 in zip(s1, s2):
if c1 != c2:
if ok:
return False
else:
ok = True
return ok
```

I do not checked for length difference because you said the two strings are equal, but for a more general approach I would add it.

If you need the position of the different character:

```
def match(s1, s2):
pos = -1
for i, (c1, c2) in enumerate(zip(s1, s2)):
if c1 != c2:
if pos != -1:
return -1
else:
pos = i
return pos
```

These are benchmarks performed with timeit, tested with match("0-001", "0-101"). I translated all solutions to py3 and removed length test.

- your solution: 5.12
- Martijn Pieters' solution: 4.92
- enrico.bacis' and lakesh's solution: 5.51
- my solution: 2.42

Tests with a longer string:

Martijn Pieters' solution:

```
timeit.timeit('match("0-0016ub5j2oi06u30tj30g6790v3nug[hoyj39867i6gy9thvb05y4b896y3n098vty98thn98qg5y4n8ygnqp", "0-0016ub5j2oi06u30tj30g6790v3gug[hoyj39867i6gy9thvb05y4b896y3n098vty98thn98qg5y4n8ygnqp")', setup="""
def match(s1, s2):
combo = zip(s1, s2)
return any(c1 != c2 for c1, c2 in combo) and all(c1 == c2 for c1, c2 in combo)
""")
```

result: 32.82

My solution:

```
timeit.timeit('match("0-0016ub5j2oi06u30tj30g6790v3nug[hoyj39867i6gy9thvb05y4b896y3n098vty98thn98qg5y4n8ygnqp", "0-0016ub5j2oi06u30tj30g6790v3gug[hoyj39867i6gy9thvb05y4b896y3n098vty98thn98qg5y4n8ygnqp")', setup="""
def match(s1, s2):
ok = False
for c1, c2 in zip(s1, s2):
if c1 != c2:
if ok:
return False
else:
ok = True
return ok
""")
```

Result: 20.21

Recommended from our users: **Dynamic Network Monitoring from WhatsUp Gold from IPSwitch**. ** Free Download**