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:
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
# 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 codelst = [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) |
[0, 1, 0, 0, 2, 2, 0, 0, 0, 3, 3, 3]
Time Complexity: O(n*n), where n is the length of the list test_list
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the res list
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
# 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 codelst = [0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1]print(transform(lst)) |
[0, 1, 0, 0, 2, 2, 0, 0, 0, 3, 3, 3]
Time Complexity: O(n*n), where n is the number of elements in the list
Auxiliary Space: O(n), where n is the number of elements in the list
