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Python | Program to print duplicates from a list of integers

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  • Difficulty Level : Easy
  • Last Updated : 19 Oct, 2022
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Given a list of integers with duplicate elements in it. The task is to generate another list, which contains only the duplicate elements. In simple words, the new list should contain elements that appear as more than one.

Examples:

Input : list = [10, 20, 30, 20, 20, 30, 40, 50, -20, 60, 60, -20, -20]
Output : output_list = [20, 30, -20, 60]
Input :  list = [-1, 1, -1, 8]
Output : output_list = [-1]

Method 1: Using the Brute Force approach

Python3




# Python program to print
# duplicates from a list
# of integers
def Repeat(x):
    _size = len(x)
    repeated = []
    for i in range(_size):
        k = i + 1
        for j in range(k, _size):
            if x[i] == x[j] and x[i] not in repeated:
                repeated.append(x[i])
    return repeated
 
# Driver Code
list1 = [10, 20, 30, 20, 20, 30, 40,
         50, -20, 60, 60, -20, -20]
print (Repeat(list1))
     
# This code is contributed
# by Sandeep_anand


Output

[20, 30, -20, 60]

Method 2: Using a single for loop

Python3




# Python program to print duplicates from
# a list of integers
lis = [1, 2, 1, 2, 3, 4, 5, 1, 1, 2, 5, 6, 7, 8, 9, 9]
 
uniqueList = []
duplicateList = []
 
for i in lis:
    if i not in uniqueList:
        uniqueList.append(i)
    elif i not in duplicateList:
        duplicateList.append(i)
 
print(duplicateList)


Output

[1, 2, 5, 9]

Method 3: Using Counter() function from collection module

Python3




from collections import Counter
 
l1 = [1,2,1,2,3,4,5,1,1,2,5,6,7,8,9,9]
d = Counter(l1)
print(d)
 
new_list = list([item for item in d if d[item]>1])
print(new_list)


Output

Counter({1: 4, 2: 3, 5: 2, 9: 2, 3: 1, 4: 1, 6: 1, 7: 1, 8: 1})
[1, 2, 5, 9]

Method 4: Using count() method

Python3




# program to print duplicate numbers in a given list
# provided input
list = [1, 2, 1, 2, 3, 4, 5, 1, 1, 2, 5, 6, 7, 8, 9, 9]
 
new = []  # defining output list
 
# condition for reviewing every
# element of given input list
for a in list:
 
     # checking the occurrence of elements
    n = list.count(a)
 
    # if the occurrence is more than
    # one we add it to the output list
    if n > 1:
 
        if new.count(a) == 0# condition to check
 
            new.append(a)
 
print(new)
 
# This code is contributed by Himanshu Khune


Output

[1, 2, 5, 9]

Method 5: Using list comprehension method

Python3




def duplicate(input_list):
    return list(set([x for x in input_list if input_list.count(x) > 1]))
 
if __name__ == '__main__':
    input_list = [1, 2, 1, 2, 3, 4, 5, 1, 1, 2, 5, 6, 7, 8, 9, 9]
    print(duplicate(input_list))
 
# This code is contributed by saikot


Output

[1, 2, 5, 9]

Method 6: Using list-dictionary approach (without any inbuild count function)

Python3




def duplicate(input_list):
    new_dict, new_list = {}, []
 
    for i in input_list:
        if not i in new_dict:
            new_dict[i] = 1
        else:
            new_dict[i] += 1
 
    for key, values in new_dict.items():
        if values > 1:
            new_list.append(key)
 
    return new_list
 
if __name__ == '__main__':
    input_list = [1, 2, 1, 2, 3, 4, 5, 1, 1, 2, 5, 6, 7, 8, 9, 9]
    print(duplicate(input_list))
 
# This code is contributed by saikot


Output

[1, 2, 5, 9]

Method 7: Using in, not in operators and count() method

Python3




lis = [1, 2, 1, 2, 3, 4, 5, 1, 1, 2, 5, 6, 7, 8, 9, 9]
x = []
y = []
for i in lis:
    if i not in x:
        x.append(i)
for i in x:
    if lis.count(i) > 1:
        y.append(i)
print(y)


Output

[1, 2, 5, 9]

Method 8: Using enumerate function

Python3




input_list = [1, 2, 1, 2, 3, 4, 5, 1, 1, 2, 5, 6, 7, 8, 9, 9]
print(list(set([x for i,x in enumerate(input_list) if input_list.count(x) > 1])))


Output

[1, 2, 5, 9]

Time Complexity: O(n)

Auxiliary Space: O(1)


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