Python | Check if any element occurs n times in given list
Last Updated :
08 May, 2023
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")
OutputThere 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))
OutputThere 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")
OutputYes, 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")
OutputYes, 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.
OutputYes, 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
OutputYes, 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
- Convert the list to a unique value list using set(), and list().
- 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")
OutputThere are 2 occurrences of 9 in [9, 8, 7, 6, 5, 9, 2]
Time Complexity: O(N)
Auxiliary Space: O(1)
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