Given strings list, perform sort by Maximum Character in String.
Input : test_list = [“neveropen”, “is”, “best”, “cs”]
Output : [“neveropen”, “is”, “cs”, “best”]
Explanation : s = s = s < t, sorted by maximum character.Input : test_list = [“apple”, “is”, “fruit”]
Output : [“apple”, “is”, “fruit”]
Explanation : p < s < t, hence order retained after sorting by max. character.
Method #1 : Using sort() + max()
In this, sorting is performed using sort() and max() is used to get maximum character from Strings.
Python3
# Python3 code to demonstrate working of# Sort Strings by Maximum Character# Using sort() + max()# get maximum character fnc.def get_max(sub): # returns maximum character return ord(max(sub))# initializing listtest_list = ["neveropen", "is", "best", "for", "cs"]# printing original listsprint("The original list is : " + str(test_list))# performing sortingtest_list.sort(key=get_max)# printing resultprint("Sorted List : " + str(test_list)) |
The original list is : ['neveropen', 'is', 'best', 'for', 'cs'] Sorted List : ['for', 'neveropen', 'is', 'cs', 'best']
Method #2 : Using sorted() + lambda + max()
In this, we perform task of sorting using sorted(), lambda and max() are used to input logic of getting maximum character.
Python3
# Python3 code to demonstrate working of# Sort Strings by Maximum Character# Using sorted() + lambda + max()# initializing listtest_list = ["neveropen", "is", "best", "for", "cs"]# printing original listsprint("The original list is : " + str(test_list))# performing sorting using sorted()# lambda function provides logicres = sorted(test_list, key=lambda sub: ord(max(sub)))# printing resultprint("Sorted List : " + str(res)) |
The original list is : ['neveropen', 'is', 'best', 'for', 'cs'] Sorted List : ['for', 'neveropen', 'is', 'cs', 'best']
Time Complexity: O(n) -> built-in functions like max takes O(n)
Auxiliary Space: O(n)
Method 3: Using heapq.nlargest():
In this approach, we can use the heapq.nlargest() function to find the n largest elements of a list based on a given key function. We can pass k=len(test_list) to get all the elements of the list, and use a lambda function to return the maximum ASCII value of each string.
Python3
import heapqtest_list = ["neveropen", "is", "best", "cs"]sorted_list = heapq.nlargest(len(test_list), test_list, key=lambda s: max(map(ord, s)))print(sorted_list) # Output: ['neveropen', 'is', 'cs', 'best'] |
['best', 'neveropen', 'is', 'cs']
Time Complexity: O(nmlog(k)), where n is the number of strings in the list, m is the length of the longest string, and k is the number of largest elements to be found. In this case, k is equal to len(test_list), so the time complexity is O(nmlog(n)).
Auxiliary Space: O(k), where k is the number of largest elements to be found. In this case, k is equal to len(test_list), so the space complexity is O(n).
Method #4: Using a custom function with list comprehension
Python3
# Python3 code to demonstrate working of# Sort Strings by Maximum Character# Using a custom function with list comprehension# initializing listtest_list = ["neveropen", "is", "best", "for", "cs"]# defining custom function to get maximum character ASCII valuedef max_char_ascii(s): return ord(max(s))# using list comprehension to create list of tupleslst = [(s, max_char_ascii(s)) for s in test_list]# sorting list of tuples based on second element (maximum character ASCII value)res = sorted(lst, key=lambda x: x[1])# extracting first element of each tuple (original string)res = [t[0] for t in res]# printing resultprint("Sorted List : " + str(res)) |
Sorted List : ['for', 'neveropen', 'is', 'cs', 'best']
Time complexity: O(n log n) (due to the use of the sorted() function)
Auxiliary space: O(n) (for the lst list of tuples)
