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Python program to Mark duplicate elements in string

Last Updated : 23 Jul, 2025
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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 list
test_list = ["gfg", "is", "best", "gfg", 
             "best", "for", "all", "gfg"]
             
# printing original list
print("The original list is : " + str(test_list))

# getting count till current using count() and slicing
res = [val + str(test_list[:idx].count(val) + 1) if test_list.count(val) > 1 else val for idx,
       val in enumerate(test_list)]

# printing result
print("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 list
test_list = ["gfg", "is", "best", "gfg", 
             "best", "for", "all", "gfg"]
             
# printing original list
print("The original list is : " + str(test_list))

# getting count till current using count() and slicing
res = 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 result
print("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 elements
import operator as op
# initializing list
test_list = ["gfg", "is", "best", "gfg",
             "best", "for", "all", "gfg"]

# printing original list
print("The original list is : " + str(test_list))

# getting count till current using count() and slicing
res = [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 result
print("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)


M

manjeet_04
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