# Python – Sort Matrix by Number of elements greater than its previous element

Given a Matrix, sort by occurrences where next element is greater than current. Compute the count of i < i + 1 in each list, sort each row by count of each of this condition in each row.

Input : test_list = [[4, 6, 2, 9, 10], [5, 3, 2, 5], [2, 4, 5, 6, 7, 7], [6, 3, 2]]
Output : [[6, 3, 2], [5, 3, 2, 5], [4, 6, 2, 9, 10], [2, 4, 5, 6, 7, 7]]
Explanation : for [4, 6, 2, 9, 10], the count is 3 as 6>=4, 9>=2 and 10>=9, similarly for [5, 3, 2, 5], [2, 4, 5, 6,  7, 7], [6, 3, 2] counts are 1,4 and 0 respectively. As, 0<1<3<4 so the order of rows is [6, 3, 2], [5, 3, 2, 5], [4, 6, 2, 9, 10], [2, 4, 5, 6, 7, 7]

Input : test_list = [[5, 3, 2, 5], [2, 4, 5, 6, 7, 7], [6, 3, 2]]
Output : [[6, 3, 2], [5, 3, 2, 5], [2, 4, 5, 6, 7, 7]]
Explanation : 0 < 1 < 4, is the greater next greater elements count. No next element is greater in 1st list.

Method #1 : Using sort() + len()

In this, we perform task of sorting using sort() and call external function as the key to solve problem of counting elements with next element greater. The size is computed using len().

## Python3

 `# Python3 code to demonstrate working of` `# Sort Matrix by Next Greater Frequency` `# Using sort() + len()`     `# getting frequency of next greater` `def` `get_greater_freq(row):`   `    ``# getting length` `    ``return` `len``([row[idx] ``for` `idx ``in` `range``(``0``, ``len``(row) ``-` `1``) ``if` `row[idx] < row[idx ``+` `1``]])`     `# initializing list` `test_list ``=` `[[``4``, ``6``, ``2``, ``9``, ``10``], [``5``, ``3``, ``2``, ``5``], [``2``, ``4``, ``5``, ``6``, ``7``, ``7``], [``6``, ``3``, ``2``]]`   `# printing original list` `print``(``"The original list is : "` `+` `str``(test_list))`   `# inplace sorting` `test_list.sort(key``=``get_greater_freq)`   `# printing result` `print``(``"Sorted rows : "` `+` `str``(test_list))`

Output:

The original list is : [[4, 6, 2, 9, 10], [5, 3, 2, 5], [2, 4, 5, 6, 7, 7], [6, 3, 2]]
Sorted rows : [[6, 3, 2], [5, 3, 2, 5], [4, 6, 2, 9, 10], [2, 4, 5, 6, 7, 7]]

Time Complexity: O(nlogn+mlogm)
Auxiliary Space: O(1)

Method #2 : Using sorted() + len() + lambda

In this, we perform task of sorting using sorted(), lambda and len() are used for creating one-liner functionality to perform sorting o the basis of number of elements greater than their previous element.

## Python3

 `# Python3 code to demonstrate working of` `# Sort Matrix by Next Greater Frequency` `# Using sorted() + len() + lambda`   `# initializing list` `test_list ``=` `[[``4``, ``6``, ``2``, ``9``, ``10``], [``5``, ``3``, ``2``, ``5``], [``2``, ``4``, ``5``, ``6``, ``7``, ``7``], [``6``, ``3``, ``2``]]`   `# printing original list` `print``(``"The original list is : "` `+` `str``(test_list))`   `# performing one-liner sorting` `# avoiding external fnc. call` `res ``=` `sorted``(test_list, key``=``lambda` `row: ``len``(` `    ``[row[idx] ``for` `idx ``in` `range``(``0``, ``len``(row) ``-` `1``) ``if` `row[idx] < row[idx ``+` `1``]]))`   `# printing result` `print``(``"Sorted rows : "` `+` `str``(res))`

Output:

The original list is : [[4, 6, 2, 9, 10], [5, 3, 2, 5], [2, 4, 5, 6, 7, 7], [6, 3, 2]]
Sorted rows : [[6, 3, 2], [5, 3, 2, 5], [4, 6, 2, 9, 10], [2, 4, 5, 6, 7, 7]]

Time Complexity: O(n*logn), where n is the length of the input list. This is because we’re using the built-in sorted() function which has a time complexity of O(nlogn) in the worst case.
Auxiliary Space: O(n), as we’re using additional space other than the input list itself.

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