Python | Program to print duplicates from a list of integers

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]

Time complexity: O(n^2) where n is the length of the input list
Auxiliary space: O(k) where k is the number of duplicates in the input list.

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]

Time Complexity: O(n)
Auxiliary Space: O(n)

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]

Time Complexity: O(n)
Auxiliary Space: O(n)

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)

Method 9: Using operator.countOf() method

Python3

import operator as op
# 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 = op.countOf(list, a)

    # if the occurrence is more than
    # one we add it to the output list
    if n > 1:

        if op.countOf(new, a) == 0:  # condition to check

            new.append(a)

print(new)

# This code is contributed by vikkycirus

Output

[1, 2, 5, 9]

Time Complexity: O(n)
Auxiliary Space: O(1)

Method 10 : Using operator.countOf(),in and not in operators

Python3

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

Output

[1, 2, 5, 9]

Time Complexity: O(n)
Auxiliary Space: O(n)

Method 11 : Using numpy:

Algorithm:

1. Initialize the list l1 with some values.
2. Import the numpy library.
3. Use numpy.unique() to get the unique values and their counts in the list l1.
4. Use numpy.where() to get the indices of the elements in the counts array that are greater than 1.
5. Use these indices to get the corresponding elements from the unique array.

Python3

import numpy as np

l1 = [1,2,1,2,3,4,5,1,1,2,5,6,7,8,9,9]

unique, counts = np.unique(l1, return_counts=True)
new_list = unique[np.where(counts > 1)]

print(new_list)
#This code is contributed by Rayudu

Output:
[1 2 5 9]

Time Complexity:

numpy.unique() has a time complexity of O(nlogn) due to the sorting step, where n is the length of the input array.
numpy.where() has a time complexity of O(n), where n is the length of the input array.
Indexing an array takes constant time O(1).
Therefore, the time complexity of this numpy code is O(nlogn), dominated by numpy.unique().

Auxiliary Space:

The space complexity of this numpy code is O(n), since we are storing the input list l1 and the output list new_list.
numpy.unique() and numpy.where() internally create intermediate arrays, but they are not included in the space complexity of this code because they are temporary and not stored in memory.