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  1. PO
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    <p>My quick attempt:</p> <pre><code>def count_subseqs(string, subseq): string = [c for c in string if c in subseq] count = i = 0 for c in string: if c == subseq[0]: pos = 1 for c2 in string[i+1:]: if c2 == subseq[pos]: pos += 1 if pos == len(subseq): count += 1 break i += 1 return count print count_subseqs(string='3141592653', subseq='123') </code></pre> <p><strong>Edit:</strong> This one should be correct also if <code>1223 == 2</code> and more complicated cases:</p> <pre><code>def count_subseqs(string, subseq): string = [c for c in string if c in subseq] i = 0 seqs = [] for c in string: if c == subseq[0]: pos = 1 seq = [1] for c2 in string[i + 1:]: if pos &gt; len(subseq): break if pos &lt; len(subseq) and c2 == subseq[pos]: try: seq[pos] += 1 except IndexError: seq.append(1) pos += 1 elif pos &gt; 1 and c2 == subseq[pos - 1]: seq[pos - 1] += 1 if len(seq) == len(subseq): seqs.append(seq) i += 1 return sum(reduce(lambda x, y: x * y, seq) for seq in seqs) assert count_subseqs(string='12', subseq='123') == 0 assert count_subseqs(string='1002', subseq='123') == 0 assert count_subseqs(string='0123', subseq='123') == 1 assert count_subseqs(string='0123', subseq='1230') == 0 assert count_subseqs(string='1223', subseq='123') == 2 assert count_subseqs(string='12223', subseq='123') == 3 assert count_subseqs(string='121323', subseq='123') == 3 assert count_subseqs(string='12233', subseq='123') == 4 assert count_subseqs(string='0123134', subseq='1234') == 2 assert count_subseqs(string='1221323', subseq='123') == 5 </code></pre>
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    1. This table or related slice is empty.
    1. POFind the number of occurrences of a subsequence in a string
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    1. USJakub
    1. USJakub
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      1. VTUndeletion
    1. COHm, it's probably more complicated than that. It would be good to see the original assignment. If, for example, this counts for two `1223` then my function is wrong.
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      1. USJakub
    2. COI guess `1223` counts for two :(
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      1. USunkulunkulu
    3. COI've added another variant which should work correctly even if `1223` counts for two, `12233` for four, etc.
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      1. USJakub
 

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