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    <p>There are 4 things I'd like to mention. Here's some code and its output:</p> <pre><code>def srk_func(words): current = [] lastc = [] for x in words: if len(current) == 0: current.append(int(x)) elif len(current) == 1: if current[0] &lt; int(x): current.append(int(x)) else: if len(current) &gt;= len(lastc): lastc = current current[:] = [] current.append(int(x)) elif len(current) &gt;= 2: if current[-1] &lt; int(x): current.append(int(x)) else: if len(current) &gt;= len(lastc): lastc = current elif len(current) &lt; len(lastc): current[:] = [] current[:] = [] current.append(int(x)) return lastc def jm_func(words): current = [] lastc = [] for w in words: x = int(w) if not current: # this happens only on the first element current = [x] continue if x &gt; current[-1]: current.append(x) else: # no increase, so current is complete if len(current) &gt;= len(lastc): lastc = current current = [x] # end of input, so current is complete if len(current) &gt;= len(lastc): lastc = current return lastc tests = """\ 1 1 5 5 1 1 5 7 7 5 1 1 5 7 0 1 5 7 0 3 1 5 7 0 2 4 6 8 1 3 5 7 9 11 0 2 """ for test in tests.splitlines(): wds = test.split() print wds print srk_func(wds) print jm_func(wds) print 8&lt;-------------------------------------------------- ['1'] [] [1] ['1', '5'] [] [1, 5] ['5', '1'] [1] [1] ['1', '5', '7'] [] [1, 5, 7] ['7', '5', '1'] [1] [1] ['1', '5', '7', '0'] [0] [1, 5, 7] ['1', '5', '7', '0', '3'] [0, 3] [1, 5, 7] ['1', '5', '7', '0', '2', '4', '6', '8'] [0, 2, 4, 6, 8] [0, 2, 4, 6, 8] ['1', '3', '5', '7', '9', '11', '0', '2'] [0, 2] [1, 3, 5, 7, 9, 11] [] [] [] </code></pre> <p>Topic 1: Test your code.</p> <p>Topic 2: Redundancy: Your code for <code>len(current) == 1</code> is functionally identical to your code for len >= 2, and the latter is bloated by having the following unnecessary two lines:</p> <pre><code> elif len(current) &lt; len(lastc): current[:] = [] </code></pre> <p>You can combine the two cases; see my version.</p> <p>Topic 3: It often happens in this kind of algorithm where you are processing input and keeping some "state" (in this case, current and lastc) that you can't immediately pack up and go home when you reach the end of the input; you need to do something with that state.</p> <p>Topic 4: This is going to get technical, but it's a common trap for new Python players. </p> <pre><code>&gt;&gt;&gt; current = [1, 2, 3, 4, 5] &gt;&gt;&gt; lastc = current # lastc and current refer to THE SAME LIST; no copying! &gt;&gt;&gt; print current [1, 2, 3, 4, 5] &gt;&gt;&gt; print lastc [1, 2, 3, 4, 5] &gt;&gt;&gt; current[:] = [] # The list to which current refers is cleared &gt;&gt;&gt; print current [] &gt;&gt;&gt; print lastc # lastc refers to the same list [] </code></pre> <p>It's better to just assign the name <code>current</code> to a new list; see my code.</p> <p>Topic 5 (bonus extra): Consider using 4-space indentation, not 8. Causing both the horizontal and vertical scroll-bars to appear in an SO question or answer == FAIL :-)</p>
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    1. USJohn Machin
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    1. COThanks for your response John. I didn't realize when I had posted the original code that I copied from PUTTY. Thank you for the tips, however - the `len(current) == 1` was put in because when I would try to reference the last number in the list, with `[-1]` it was giving me an error. Also, comparing `len(current) > len(lastc)` was incorrectly assigning the list of current to last. After inserting multiple print statements, I found that `long_run = max(len(currnet),long_run)` worked much easier. Topic 4 - Wow John, if you hadn't explicitly put that - I would have never known.
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