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# Python | Increment 1’s in list based on pattern

• Last Updated : 30 Jul, 2019

Given a list of binary numbers 0 and 1, Write a Python program to transform the list in such a way that whenever 1 appears after the occurrence of a sequence of 0’s, increment it by n+1, where ‘n’ is the last increment.

Examples:

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```Input : [0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1]
Output : [0, 1, 0, 0, 2, 2, 0, 0, 0, 3, 3, 3]

Input : [1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1]
Output : [1, 0, 2, 0, 0, 0, 3, 3, 0, 0, 4, 4, 4, 0, 5, 0, 6]
```

Approach #1 : Naive Approach

This is a naive approach to the given problem. It uses two variable ‘previous’ and ‘grp’ to store previously incremented number and to store the number of 1’s in a group. Now, using a for loop, increment 1’s accordingly.

 `# Python3 program to increment 1's in ``# list based on pattern `` ` `def` `transform(lst):``     ` `    ``previous ``=` `0``    ``grp ``=` `0``    ``for` `elem ``in` `lst:``        ``if` `elem ``and` `not` `previous:``             ``grp ``+``=` `1``        ``previous ``=` `elem``        ``yield` `(grp ``if` `elem ``else` `0``)``     ` `# Driver code``lst ``=` `[``0``, ``1``, ``0``, ``0``, ``1``, ``1``, ``0``, ``0``, ``0``, ``1``, ``1``, ``1``]``x ``=` `(transform(lst))``res ``=` `[]``for` `i ``in` `range``(``0``, ``len``(lst)):``    ``res.append(``next``(x))``print``(res)`
Output:
```[0, 1, 0, 0, 2, 2, 0, 0, 0, 3, 3, 3]
```

Approach #2 : Using count, chain and groupby from itertools module.
This is an efficient and more pythonic approach towards the given problem where we use count, chain and groupby from itertools module.

 `# Python3 program to increment 1's in ``# list based on pattern ``from` `itertools ``import` `*` ` ` `def` `transform(lst):``     ` `    ``c ``=` `count(``1``)``    ``return` `list``(chain(``*``[``list``(g) ``if` `k !``=` `1` `else` `[``next``(c)]``*``len``(``list``(g)) ``    ``for` `k, g ``in` `groupby(lst)]))``     ` `# Driver code``lst ``=` `[``0``, ``1``, ``0``, ``0``, ``1``, ``1``, ``0``, ``0``, ``0``, ``1``, ``1``, ``1``]``print``(transform(lst))`
Output:
```[0, 1, 0, 0, 2, 2, 0, 0, 0, 3, 3, 3]
```

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