Given a list, the task is to write a Python program to mark the duplicate occurrence of elements with progressive occurrence number.
Input : test_list = [‘gfg’, ‘is’, ‘best’, ‘gfg’, ‘best’, ‘for’, ‘all’, ‘gfg’]
Output : [‘gfg1’, ‘is’, ‘best1’, ‘gfg2’, ‘best2’, ‘for’, ‘all’, ‘gfg3’]
Explanation : gfg’s all occurrence are marked as it have multiple repetitions(3).
Input : test_list = [‘gfg’, ‘is’, ‘best’, ‘best’, ‘for’, ‘all’]
Output : [‘gfg’, ‘is’, ‘best1’, ‘best2’, ‘for’, ‘all’]
Explanation : best’s all occurrence are marked as it have multiple repetitions(2).
Method 1: Using count() + enumerate() + list comprehension + slicing
In this, in order to get the duplicate count, the list is sliced to current element index, and count of occurrence of that element till current index is computed using count() and append.
Python3
# Python3 code to demonstrate working of# Mark duplicate elements# Using count() + enumerate() + list comprehension + slicing # initializing listtest_list = ["gfg", "is", "best", "gfg", "best", "for", "all", "gfg"] # printing original listprint("The original list is : " + str(test_list))# getting count till current using count() and slicingres = [val + str(test_list[:idx].count(val) + 1) if test_list.count(val) > 1 else val for idx, val in enumerate(test_list)]# printing resultprint("Duplicates marked List : " + str(res)) |
Output:
The original list is : [‘gfg’, ‘is’, ‘best’, ‘gfg’, ‘best’, ‘for’, ‘all’, ‘gfg’]
Duplicates marked List : [‘gfg1’, ‘is’, ‘best1’, ‘gfg2’, ‘best2’, ‘for’, ‘all’, ‘gfg3’]
Time Complexity: O(n)
Auxiliary Space : O(n)
Method 2: Using map() + count() + lambda
Similar to the above method, the only difference being map() is used to get a function using lambda to extend to whole list elements.
Python3
# Python3 code to demonstrate working of# Mark duplicate elements# Using map() + count() + lambda# initializing listtest_list = ["gfg", "is", "best", "gfg", "best", "for", "all", "gfg"] # printing original listprint("The original list is : " + str(test_list))# getting count till current using count() and slicingres = list(map(lambda ele: ele[1] + str(test_list[ : ele[0]].count(ele[1]) + 1) if test_list.count(ele[1]) > 1 else ele[1], enumerate(test_list)))# printing resultprint("Duplicates marked List : " + str(res)) |
Output:
The original list is : [‘gfg’, ‘is’, ‘best’, ‘gfg’, ‘best’, ‘for’, ‘all’, ‘gfg’]
Duplicates marked List : [‘gfg1’, ‘is’, ‘best1’, ‘gfg2’, ‘best2’, ‘for’, ‘all’, ‘gfg3’]
The Time and Space Complexity of all the methods is :
Time Complexity: O(n)
Space Complexity: O(n)
Method #3: Using operator.countOf() method
Python3
# Python3 code to demonstrate working of# Mark duplicate elementsimport operator as op# initializing listtest_list = ["gfg", "is", "best", "gfg", "best", "for", "all", "gfg"]# printing original listprint("The original list is : " + str(test_list))# getting count till current using count() and slicingres = [val + str(op.countOf(test_list[:idx], val) + 1) if op.countOf(test_list, val) > 1 else val for idx, val in enumerate(test_list)]# printing resultprint("Duplicates marked List : " + str(res)) |
The original list is : ['gfg', 'is', 'best', 'gfg', 'best', 'for', 'all', 'gfg'] Duplicates marked List : ['gfg1', 'is', 'best1', 'gfg2', 'best2', 'for', 'all', 'gfg3']
Time Complexity: O(n)
Auxiliary Space: O(n)
