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 greaterdef 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 listtest_list = [[4, 6, 2, 9, 10], [5, 3, 2, 5], [2, 4, 5, 6, 7, 7], [6, 3, 2]]# printing original listprint("The original list is : " + str(test_list))# inplace sortingtest_list.sort(key=get_greater_freq)# printing resultprint("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 listtest_list = [[4, 6, 2, 9, 10], [5, 3, 2, 5], [2, 4, 5, 6, 7, 7], [6, 3, 2]]# printing original listprint("The original list is : " + str(test_list))# performing one-liner sorting# avoiding external fnc. callres = sorted(test_list, key=lambda row: len( [row[idx] for idx in range(0, len(row) - 1) if row[idx] < row[idx + 1]]))# printing resultprint("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.
