# Python – Test if all rows contain any common element with other Matrix

• Difficulty Level : Basic
• Last Updated : 15 Sep, 2021

Given two Matrices, check if all rows contain at least one element common with the same index row of other Matrix.

Input : test_list1 = [[5, 6, 1], [2, 4], [9, 3, 5]], test_list2 = [[9, 1, 2], [9, 8, 2], [3, 7, 10]]
Output : True
Explanation : 1, 2 and 3 are common elements in rows.

Input : test_list1 = [[5, 6, 1], [2, 4], [9, 2, 6]], test_list2 = [[9, 1, 2], [9, 8, 2], [3, 7, 6]]
Output : True
Explanation : 1, 2 and 6 are common elements in rows.

Method #1 : Using loop + in operator

In this, we iterate each row of matrix, and check for any element matching in other list using in operator, if any element is found, we mark it true, if all rows are true, True is returned.

## Python3

 `# Python3 code to demonstrate working of``# Test if all rows contain any common element with other Matrix``# Using loop "+" in operator` `# initializing lists``test_list1 ``=` `[[``5``, ``6``, ``1``], [``2``, ``4``], [``9``, ``3``, ``5``]]``test_list2 ``=` `[[``9``, ``1``, ``2``], [``9``, ``8``, ``2``], [``3``, ``7``, ``10``]]` `# printing original lists``print``(``"The original list 1 is : "` `+` `str``(test_list1))``print``(``"The original list 2 is : "` `+` `str``(test_list2))` `res ``=` `True``for` `idx ``in` `range``(``0``, ``len``(test_list1)):``    ` `    ``temp ``=` `False``    ` `    ``# checking for any common element in 2nd list``    ``for` `ele ``in` `test_list1[idx]:``        ``if` `ele ``in` `test_list2[idx]:``            ``temp ``=` `True``            ``break``        ` `    ``# if any element not found, Result is false   ``    ``if` `not` `temp :``        ``res ``=` `False``        ``break` `# printing result``print``(``"All row contain common elements with other Matrix : "` `+` `str``(res))`

Output:

```The original list 1 is : [[5, 6, 1], [2, 4], [9, 3, 5]]
The original list 2 is : [[9, 1, 2], [9, 8, 2], [3, 7, 10]]
All row contain common elements with other Matrix : True```

Method #2 : Using any() + loop

In this, one nested loop is avoided and the result is computed using any() for getting any common element in the 2nd Matrix.

## Python3

 `# Python3 code to demonstrate working of``# Test if all rows contain any common element with other Matrix``# Using loop any() "+" loop` `# initializing lists``test_list1 ``=` `[[``5``, ``6``, ``1``], [``2``, ``4``], [``9``, ``3``, ``5``]]``test_list2 ``=` `[[``9``, ``1``, ``2``], [``9``, ``8``, ``2``], [``3``, ``7``, ``10``]]` `# printing original lists``print``(``"The original list 1 is : "` `+` `str``(test_list1))``print``(``"The original list 2 is : "` `+` `str``(test_list2))` `res ``=` `True``for` `idx ``in` `range``(``0``, ``len``(test_list1)):``    ` `    ``# checking for common element in list 2 in same index``    ``temp ``=` `any``(ele ``in` `test_list2[idx] ``for` `ele ``in` `test_list1[idx])``        ` `    ``# if any element not found, Result is false   ``    ``if` `not` `temp :``        ``res ``=` `False``        ``break` `# printing result``print``(``"All row contain common elements with other Matrix : "` `+` `str``(res))`

Output:

```The original list 1 is : [[5, 6, 1], [2, 4], [9, 3, 5]]
The original list 2 is : [[9, 1, 2], [9, 8, 2], [3, 7, 10]]
All row contain common elements with other Matrix : True```

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