Python control flow with in keyword unexpected behaviour

Question

I am trying to print "YES" or "NO" (quotes for clarity) depending on whether any permutation of the pattern exists in the text string.

from itertools import permutations


i = int(input())

for j in range(i):
    m = input()
    n = input()

    lst = []
    for k in permutations(m):
        lst.append("".join(k))

    for l in lst:
        if l in n:
            print("YES")
            break
        else:
            print("NO")
            break

    print(lst)

This is the sample input:

3
hack
indiahacks
code
eddy
coder
iamredoc

This is the expected output

YES
NO
YES

For first two inputs, output is coming correctly and the third output is the one which fails.

/home/ajay/.pyenv/versions/3.4.1/bin/python3.4 /home/ajay/PycharmProjects/LearnPython/test.py
1
coder
iamredoc
NO
['coder', 'codre', 'coedr', 'coerd', 'corde', 'cored', 'cdoer', 'cdore', 'cdeor', 'cdero', 'cdroe', 'cdreo', 'ceodr', 'ceord', 'cedor', 'cedro', 'cerod', 'cerdo', 'crode', 'croed', 'crdoe', 'crdeo', 'creod', 'credo', 'ocder', 'ocdre', 'ocedr', 'ocerd', 'ocrde', 'ocred', 'odcer', 'odcre', 'odecr', 'oderc', 'odrce', 'odrec', 'oecdr', 'oecrd', 'oedcr', 'oedrc', 'oercd', 'oerdc', 'orcde', 'orced', 'ordce', 'ordec', 'orecd', 'oredc', 'dcoer', 'dcore', 'dceor', 'dcero', 'dcroe', 'dcreo', 'docer', 'docre', 'doecr', 'doerc', 'dorce', 'dorec', 'decor', 'decro', 'deocr', 'deorc', 'derco', 'deroc', 'drcoe', 'drceo', 'droce', 'droec', 'dreco', 'dreoc', 'ecodr', 'ecord', 'ecdor', 'ecdro', 'ecrod', 'ecrdo', 'eocdr', 'eocrd', 'eodcr', 'eodrc', 'eorcd', 'eordc', 'edcor', 'edcro', 'edocr', 'edorc', 'edrco', 'edroc', 'ercod', 'ercdo', 'erocd', 'erodc', 'erdco', 'erdoc', 'rcode', 'rcoed', 'rcdoe', 'rcdeo', 'rceod', 'rcedo', 'rocde', 'roced', 'rodce', 'rodec', 'roecd', 'roedc', 'rdcoe', 'rdceo', 'rdoce', 'rdoec', 'rdeco', 'rdeoc', 'recod', 'recdo', 'reocd', 'reodc', 'redco', 'redoc']

Process finished with exit code 0

You can see that redoc is present in the list of permutations but why the python in keyword is not recognizing it and thus producing NO output.

So if my answer helps, you'll add 2 points to your rep by accepting it. By the way, I did plus one this one for being interesting. — Aaron Hall
– Aaron Hall ♦, Commented Sep 15, 2014 at 13:05

Aaron Hall · Accepted Answer · 2014-09-14 14:58:06Z

4

You're breaking out of the loop over your list immediately after checking the first item in the list.

instead of:

for l in lst:
    if l in n:
        print("YES")
        break
    else...

If you must use a loop structure, you can rehabilitate your code with:

for l in lst:
    if l in n:
        print("YES")
        break
else: # this will run after (and only if) exhausting your loop.
    print("NO")

Raymond Hettinger thought this usage of else should be called nobreak and I think that's good intuition for it.

This is a usage of the else clause on loops, and is documented here.

edited Sep 14, 2014 at 14:58

answered Sep 14, 2014 at 13:43

Aaron Hall♦

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4 Comments

c_coder Over a year ago

It works thanks :) Can you explain why this particular else out of the loop works. i m not getting it exactly.

Aaron Hall Over a year ago

It's control flow that only runs if the loop ends without breaking. That means you didn't find the substring, so you want NO printed there.

ajknzhol Over a year ago

The time limit may exceed in this case. what if we want to do it within 2 sec.

Aaron Hall Over a year ago

Nothing in the question about a time limit, but if you want to guard against that, you'll have to build it into your control flow.

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