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]