Python | Check if any element occurs n times in given list

Given a list, the task is to find whether any element occurs ‘n’ times in given list of integers. It will basically check for the first element that occurs n number of times. 

Examples: 

Input: l =  [1, 2, 3, 4, 0, 4, 3, 2, 1, 2], n = 3
Output :  2

Input: l =  [1, 2, 3, 4, 0, 4, 3, 2, 1, 2, 1, 1], n = 4
Output :  1

Below are some methods to do the task in Python – 
Method 1: Using Simple Iteration and Sort 

Python3

# Python code to find occurrence of any element
# appearing 'n' times

# Input Initialisation
input = [1, 2, 3, 0, 4, 3, 4, 0, 0]

# Sort Input
input.sort()

# Constants Declaration
n = 3
prev = -1
count = 0
flag = 0

# Iterating
for item in input:
    if item == prev:
        count = count + 1
    else:
        count = 1
    prev = item

    if count == n:
        flag = 1
        print("There are {} occurrences of {} in {}".format(n, item, input))
        break

# If no element is not found.
if flag == 0:
    print("No occurrences found")

Output

There are 3 occurrences of 0 in [0, 0, 0, 1, 2, 3, 3, 4, 4]

Time complexity: O(n*logn), as sort() function is used.
Auxiliary space: O(1)

Method 2: Using Count

Python3

# Python code to find occurrence of any element
# appearing 'n' times

# Input list initialisation
input = [1, 2, 3, 4, 0, 4, 3, 4]

# Constant declaration
n = 3

# print
print("There are {} occurrences of {} in {}".format(input.count(n), n, input))

Output

There are 2 occurrences of 3 in [1, 2, 3, 4, 0, 4, 3, 4]

Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(1), as only a constant amount of extra space is used.

Method 3: Using defaultdict
We first populate item of list in a dictionary and then we find whether count of any element is equal to n. 

Python3

# Python code to find occurrence of any element
# appearing 'n' times

# importing
from collections import defaultdict

# Dictionary declaration
dic = defaultdict(int)

# Input list initialisation
Input = [9, 8, 7, 6, 5, 9, 2]

# Dictionary populate
for i in Input:
    dic[i]+= 1

# constant declaration
n = 2
flag = 0

# Finding from dictionary
for element in Input:
    if element in dic.keys() and dic[element] == n:
        print("Yes, {} has {} occurrence in {}.".format(element, n, Input))
        flag = 1
        break

# if element not found.
if flag == 0:
    print("No occurrences found")

Output

Yes, 9 has 2 occurrence in [9, 8, 7, 6, 5, 9, 2].

Method #4 : Using Counter() function

Python3

# Python code to find occurrence of any element
# appearing 'n' times

# importing
from collections import Counter

# Input list initialisation
Input = [9, 8, 7, 6, 5, 9, 2]

freq = Counter(Input)
# constant declaration
n = 2


if n in freq.values():
    for key, value in freq.items():
        if(value == n):
            print("Yes, {} has {} occurrence in {}.".format(key, n, Input))

# if element not found.
else:
    print("No occurrences found")

Output

Yes, 9 has 2 occurrence in [9, 8, 7, 6, 5, 9, 2].

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

Method#5:Using regular expressions

Python3

import re

# Input list initialisation
input_list = [9, 8, 7, 6, 5, 9, 2]

# constant declaration
n = 2

occurrences = {}
for item in input_list:
    occurrences[item] = occurrences.get(item, 0) + 1

result = [item for item, count in occurrences.items() if count == n]

if result:
     print("Yes, {} has {} occurrence in {}.".format(result[0], n, input_list))
else:
    print("No items appear {} times in the input list".format(n))
#This code is contributed by Vinay Pinjala.

Output

Yes, 9 has 2 occurrence in [9, 8, 7, 6, 5, 9, 2].

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

Method#6: Using Recursive method.

Python3

def count_occurrences(input_list, item, n):
    if not input_list:
        return False

    if input_list[0] == item:
        n -= 1

    if n == 0:
        return True

    return count_occurrences(input_list[1:], item, n)

def main():
    input_list = [9, 8, 7, 6, 5, 9, 2]
    n = 2

    for item in set(input_list):
        if count_occurrences(input_list, item, n):
            print("Yes, {} has {} occurrence in {}.".format(item, n, input_list))
            return

    print("No items appear {} times in the input list".format(n))

if __name__ == '__main__':
    main()
#this code contributed by tvsk

Output

Yes, 9 has 2 occurrence in [9, 8, 7, 6, 5, 9, 2].

Time Complexity: O(n)

Auxiliary Space: O(n)

Method #7 : Using operator.countOf() method

Approach

1. Convert the list to a unique value list using set(), and list().
2. Check the  unique element count in the original list is equal to N or not, If equal print(“There are {} occurrences of {} in {}”.format(n, i , input)), Otherwise, print(“No occurrences found”)

Python3

# Python code to find occurrence of any element
# appearing 'n' times
import operator

# Input list initialisation
input = [9, 8, 7, 6, 5, 9, 2]

# Constant declaration
n = 2

x = list(set(input))
res = False
for i in x:
    if(operator.countOf(input, i) == n):
        print("There are {} occurrences of {} in {}".format(n, i, input))
        res = True
        break

if(not res):
    print("No occurrences found")

Output

There are 2 occurrences of 9 in [9, 8, 7, 6, 5, 9, 2]

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