# Python Counter| Find duplicate rows in a binary matrix

• Last Updated : 29 Jul, 2022

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)```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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,

1. Create a dictionary using counter method which will have rows as key and it’s frequency as value.
2. Now traverse dictionary completely and print all rows which have frequency greater than 1.

Implementation:

## Python3

 `# Function to find duplicate rows in a binary matrix``from` `collections ``import` `Counter` `def` `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 program``if` `__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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