# Python | Make a list of intervals with sequential numbers

Given a list of sequential numbers, Write a Python program to convert the given list into list of intervals.

Examples:

```Input : [2, 3, 4, 5, 7, 8, 9, 11, 15, 16]
Output : [[2, 5], [7, 11], [15, 16]]

Input : [1, 2, 3, 6, 7, 8, 9, 10]
Output : [[1, 3], [6, 10]]```

Method #1 : Naive Approach First, we use the brute force approach to Convert list of sequential number into intervals. Start a loop till the length of the list. In every iteration, use another loop to check the continuity of the sequence. As soon as the sequence stop, yield the lower and higher bound of each interval.

## Python3

 `# Python3 program to Convert list of ``# sequential number into intervals` `def` `interval_extract(``list``):``    ``length ``=` `len``(``list``)``    ``i ``=` `0``    ``while` `(i< length):``        ``low ``=` `list``[i]``        ``while` `i ``=` `1``):``            ``yield` `[low, high]``        ``elif` `(high ``-` `low ``=``=` `1``):``            ``yield` `[low, ]``            ``yield` `[high, ]``        ``else``:``            ``yield` `[low, ]``        ``i ``+``=` `1` `# Driver code``l ``=`  `[``2``, ``3``, ``4``, ``5``, ``7``, ``8``, ``9``, ``11``, ``15``, ``16``]``print``( ``list``(interval_extract(l)))`

Output:
`[[2, 5], [7, 9], [11], [15, 16]]`

Time Complexity: O(n), where n is the length of the list
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the list

Method #2 : Pythonic Naive First, sort the given list. Initialize previous_number and range_start with first element. Start a loop and check if the next number is additive of the previous number, If yes, Initialize this number to previous number otherwise yield the new interval starting with range_start and ending with previous_number

## Python3

 `# Python3 program to Convert list of ``# sequential number into intervals` `def` `interval_extract(``list``):``    ``list` `=` `sorted``(``set``(``list``))``    ``range_start ``=` `previous_number ``=` `list``[``0``]` `    ``for` `number ``in` `list``[``1``:]:``        ``if` `number ``=``=` `previous_number ``+` `1``:``            ``previous_number ``=` `number``        ``else``:``            ``yield` `[range_start, previous_number]``            ``range_start ``=` `previous_number ``=` `number``    ``yield` `[range_start, previous_number]` `# Driver code``l ``=` `[``2``, ``3``, ``4``, ``5``, ``7``, ``8``, ``9``, ``11``, ``15``, ``16``]``print``( ``list``(interval_extract(l)))`

Output:
`[[2, 5], [7, 9], [11, 11], [15, 16]]`

Method #3 : Using itertools The other pythonic method is to use Python itertools. We use itertools.groupby(). Where enumerate(iterable) is taken as iterable and lambda t: t[1] – t[0]) as key function to find the sequence for intervals.

## Python3

 `# Python3 program to Convert list of ``# sequential number into intervals``import` `itertools` `def` `intervals_extract(iterable):``    ` `    ``iterable ``=` `sorted``(``set``(iterable))``    ``for` `key, group ``in` `itertools.groupby(``enumerate``(iterable),``    ``lambda` `t: t[``1``] ``-` `t[``0``]):``        ``group ``=` `list``(group)``        ``yield` `[group[``0``][``1``], group[``-``1``][``1``]]` `# Driver code``l ``=` `[``2``, ``3``, ``4``, ``5``, ``7``, ``8``, ``9``, ``11``, ``15``, ``16``]``print``( ``list``(intervals_extract(l)))`

Output:
`[[2, 5], [7, 9], [11, 11], [15, 16]]`

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