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]

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]

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]