Given a binary matrix whose elements are only 0 and 1, we need to print the rows which are duplicate of rows which are already present in the matrix.
Examples:
Input : [[1, 1, 0, 1, 0, 1],
[0, 0, 1, 0, 0, 1],
[1, 0, 1, 1, 0, 0],
[1, 1, 0, 1, 0, 1],
[0, 0, 1, 0, 0, 1],
[0, 0, 1, 0, 0, 1]]
Output : (1, 1, 0, 1, 0, 1)
(0, 0, 1, 0, 0, 1)
We have existing solution for this problem please refer Find duplicate rows in a binary matrix link. We can solve this problem very quickly in Python using Counter() method. Approach is very simple,
- Create a dictionary using counter method which will have rows as key and it’s frequency as value.
- Now traverse dictionary completely and print all rows which have frequency greater than 1.
Implementation:
Python3
# Function to find duplicate rows in a binary matrixfrom collections import Counterdef duplicate(input): # since lists are unhasable for counter method # because lists are mutable so first we will cast # each row (list) into tuple input = map(tuple,input) # now create dictionary freqDict = Counter(input) # print all rows having frequency greater than 1 for (row,freq) in freqDict.items(): if freq>1: print (row)# Driver programif __name__ == "__main__": input = [[1, 1, 0, 1, 0, 1], [0, 0, 1, 0, 0, 1], [1, 0, 1, 1, 0, 0], [1, 1, 0, 1, 0, 1], [0, 0, 1, 0, 0, 1], [0, 0, 1, 0, 0, 1]] duplicate(input) |
Output
(1, 1, 0, 1, 0, 1) (0, 0, 1, 0, 0, 1)
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