Python – Test if all elements are unique in columns in a Matrix
Last Updated :
27 Mar, 2023
Given a Matrix, test if all columns contain unique elements.
Input : test_list = [[3, 4, 5], [1, 2, 4], [4, 1, 10]]
Output : True
Explanation : 3, 1, 4; 4, 2, 1; 5, 4, 10; All elements are unique in columns.
Input : test_list = [[3, 4, 5], [3, 2, 4], [4, 1, 10]]
Output : False
Explanation : 3, 3, 4; 3 repeated twice.
Method #1 : Using loop + set() + len()
In this, we iterate for each column and test for unique elements using set size using len(), if any column is found having a size not equal to the actual list, then the result is flagged off.
Python3
test_list = [[3, 4, 5], [1, 2, 4], [4, 1, 10]]
print("The original list is : " + str(test_list))
res = True
for idx in range(len(test_list[0])):
col = [ele[idx] for ele in test_list]
if len(list(set(col))) != len(col):
res = False
break
print("Are all columns Unique : " + str(res))
|
Output:
The original list is : [[3, 4, 5], [1, 2, 4], [4, 1, 10]]
Are all columns Unique : True
Time Complexity: O(n*m)
Auxiliary Space: O(k)
Method #2 : Using all() + list comprehension
This can be solved in one-liner using all() which checks for all the columns, made using list comprehension, if all columns return True, all() returns true.
Python3
test_list = [[3, 4, 5], [1, 2, 4], [4, 1, 10]]
print("The original list is : " + str(test_list))
res = True
for idx in range(len(test_list[0])):
col = [ele[idx] for ele in test_list]
if len(list(set(col))) != len(col):
res = False
break
print("Are all columns Unique : " + str(res))
|
Output:
The original list is : [[3, 4, 5], [1, 2, 4], [4, 1, 10]]
Are all columns Unique : True
Time Complexity: O(n) where n is the number of elements in the list “test_list”. The time complexity of the all() and list comprehension is O(n)
Auxiliary Space: O(1), no extra space is required
Method #3 : Using loop + Counter() function
Python3
from collections import Counter
test_list = [[3, 4, 5], [1, 2, 4], [4, 1, 10]]
print("The original list is : " + str(test_list))
res = True
for idx in range(len(test_list[0])):
col = []
for ele in test_list:
col.append(ele[idx])
freq = Counter(col)
if(len(freq.keys()) != len(col)):
res = False
break
print("Are all columns Unique : " + str(res))
|
Output
The original list is : [[3, 4, 5], [1, 2, 4], [4, 1, 10]]
Are all columns Unique : True
Time Complexity: O(N*M)
Auxiliary Space: O(N*M)
Method #4: Using numpy:
Algorithm:
- Initialize the list to check for uniqueness in columns.
- For each column in the transposed matrix of the input list, check if the number of unique elements is the same as the length of the column.
- If any column is found to have duplicate elements, set the result to False and break the loop.
- If all columns have unique elements, set the result to True.
- Print the result.
Python3
import numpy as np
test_list = [[3, 4, 5], [1, 2, 4], [4, 1, 10]]
print("The original list is : " + str(test_list))
res = all(len(np.unique(col)) == len(col) for col in zip(*test_list))
print("Are all columns Unique : " + str(res))
|
Output:
The original list is : [[3, 4, 5], [1, 2, 4], [4, 1, 10]]
Are all columns Unique : True
Time complexity: O(nm log m)
where n is the number of rows and m is the number of columns in the input matrix. This is because we are iterating over each column of the transposed matrix and checking if the length of the unique elements in each column is equal to the length of the column. The time complexity of np.unique() function is O(m log m) where m is the number of elements in the array.
Auxiliary Space: O(m)
where m is the number of columns in the input matrix. This is because we are storing each column temporarily in the list comprehension before checking for uniqueness. Additionally, the space complexity of the np.unique() function is O(m).
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