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    <p>Using a generator function makes the task easy.</p> <pre><code>table = [(2,3),(5,6),(12,20),(21,25),(28,28),(35,48),(53,55)] def gaps_between(intervals): prec = intervals[0][1] + 1 for L,H in intervals: print '\nprec = %d (L,H) = (%d,%d)' % (prec,L,H) print 'prec = %d &lt;= L-1 = %d : %s' % (prec,L-1,prec&lt;=L) if prec&lt;=L-1: yield (prec,L-1) prec = H + 1 print 'next prec = %d' % prec holes = list(gaps_between(table)) print print 'table =',table print 'holes =',holes </code></pre> <p>In a former answer, I had used an iterator defined inside the generator.<br> To avoid this, I here above use a stratagem :<br> defining <code>first prec = first H = intervals[0][1]</code>.<br> Given the fact that <code>H&gt;=L</code> for every couple (L,H) , it leads to<br> <code>first H &gt; first L - 1</code> --> <code>first prec &gt; first L - 1</code>.<br> Hence the first test concerning the first interval is always False and the real process starts at the second interval.</p> <pre><code>prec = 3 (L,H) = (2,3) prec = 3 &lt;= L-1 = 1 : False next prec = 4 prec = 4 (L,H) = (5,6) prec = 4 &lt;= L-1 = 4 : True next prec = 7 prec = 7 (L,H) = (12,20) prec = 7 &lt;= L-1 = 11 : True next prec = 21 prec = 21 (L,H) = (21,25) prec = 21 &lt;= L-1 = 20 : True next prec = 26 prec = 26 (L,H) = (28,28) prec = 26 &lt;= L-1 = 27 : True next prec = 29 prec = 29 (L,H) = (35,48) prec = 29 &lt;= L-1 = 34 : True next prec = 49 prec = 49 (L,H) = (53,55) prec = 49 &lt;= L-1 = 52 : True next prec = 56 table = [(2, 3), (5, 6), (12, 20), (21, 25), (28, 28), (35, 48), (53, 55)] holes = [(4, 4), (7, 11), (26, 27), (29, 34), (49, 52)] </code></pre> <p>The result is correct:<br> - it gives the gap (4,4) between (2,3) and (5,6)<br> - there's no gap between (12,20) and (21,25)<br> - value 28 present in (28,28) isn't in any gap</p> <p>OP says that the intervals are non over-lapping and sorted.<br> However, the test <code>if prec&lt;=L-1</code> is mandatory, otherwise contiguous intervals give an error:<br> without this test, the result would contain<br> <code>....., (7, 11), (21, 20), (26, 27), .......</code></p> <p>. </p> <p>It happens that with this mandatory test, the following lists of intervals<br> <code>[[ 8, 9],[14, 18],[18, 32]]</code><br> <code>[[8, 9], [14, 18], [19, 20], [16, 21], [23, 32]]</code><br> that are over-lapping (not what OP says)<br> and that would give error without the test,<br> don't produce any error in fact.</p> <p>The rule for a list of intervals to give a correct list of gaps with my above code is that the intervals must be sorted along the second element.</p> <p>.</p> <p>Replacing <code>yield (prec,L-1)</code> with <code>yield range(prec,L)</code> will gives the gaps as ranges.</p> <p>Replacing <code>yield (prec,L-1)</code> with <code>holes.append((prec,L-1))</code>, for example, allows to write the code without a function.</p>
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    1. COStoring `prec` in the generator makes for a nice readable solution. You don't need the `else` clause, just move `prec = b` after the `if` block. The OP now says the pairs are sorted and non-overlapping, so you don't need to test for those cases. But I think it probably makes sense to suppress "empty holes" like (4, 4) and (22, 22) from the output if they can occur.
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    2. COAh ! you are right for the line ``prec = b``. Instrcutions ``a += 1`` and ``b += 1`` can also be removed if ``prec = b + 1``. - But for (4,4) and (22,22), it's correct: they mean gaps of only one number. - I imagined other solutions, based on ouputs directly being ranges. - Thank you for your comment
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    3. COIf you want to avoid modifying table, one way is to just set `prec = None` initially and test for it in the loop
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