Python | Check if k occurs atleast n times in a list
Last Updated :
10 May, 2023
There are many common problems that a developer or common programmer comes across in day-day coding. One such problem is finding if an element occurs more than a certain number of times in the list. Let’s discuss certain ways to deal with this problem.
Method #1 : Using sum() + list comprehension The list comprehension can be clubbed with the sum function for achieving this particular task. The sum function does the summation part and the logical case of returning true is dealt in list comprehension part.
Python3
test_list = [1, 3, 5, 5, 4, 5]
print ("The original list is : " + str(test_list))
k = 5
N = 3
res = 0
res = sum([1 for i in test_list if i == k]) >= N
if res == 1 :
res = True
else :
res = False
print ("Does %d occur atleast %d times ? :" %(k, N) + str(res))
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Output
The original list is : [1, 3, 5, 5, 4, 5]
Does 5 occur atleast 3 times ? :True
Method #2 : Using next() + islice() These both functions can be used together to perform this particular task in more efficient manner than above. The islice function would handle the summation part and next function helps with iteration logic.
Python3
from itertools import islice
test_list = [1, 3, 5, 5, 4, 5]
print ("The original list is : " + str(test_list))
k = 5
N = 3
func = (True for i in test_list if i == k)
res = next(islice(func, N-1, None), False)
print ("Does %d occur atleast %d times ? :" %(k, N) + str(res))
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Output:
The original list is : [1, 3, 5, 5, 4, 5]
Does 5 occur atleast 3 times ? : True
Method #3: Using count()
Python3
test_list = [1, 3, 5, 5, 4, 5]
print ("The original list is : " + str(test_list))
k = 5
N = 3
res = False
if(test_list.count(k)>=N):
res=True
print ("Does %d occur atleast %d times ? : " %(k, N) + str(res))
|
Output
The original list is : [1, 3, 5, 5, 4, 5]
Does 5 occur atleast 3 times ? : True
Method #4 : Using for loop
Python3
test_list = [1, 3, 5, 5, 4, 5]
print ("The original list is : " + str(test_list))
k = 5
N = 3
res = False
c=0
for i in test_list:
if(i==k):
c+=1
if(c>=N):
res=True
print ("Does %d occur atleast %d times ? : " %(k, N) + str(res))
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Output
The original list is : [1, 3, 5, 5, 4, 5]
Does 5 occur atleast 3 times ? : True
Time Complexity: O(n), where n is length of test_list.
Auxiliary Space: O(1)
Method #5 : Using operator.countOf() method
Python3
import operator as op
test_list = [1, 3, 5, 5, 4, 5]
print("The original list is : " + str(test_list))
k = 5
N = 3
res = False
if(op.countOf(test_list, k) >= N):
res = True
print("Does %d occur atleast %d times ? : " % (k, N) + str(res))
|
Output
The original list is : [1, 3, 5, 5, 4, 5]
Does 5 occur atleast 3 times ? : True
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #6: Using the itertools.groupby function
Python3
import itertools
test_list = [1, 3, 5, 5, 4, 5]
print("The original list is : " + str(test_list))
k = 5
N = 3
res = any(len(list(g)) >= N for _, g in itertools.groupby(test_list) if _ == k)
print("Does %d occur atleast %d times ? : " % (k, N) + str(res))
|
Output
The original list is : [1, 3, 5, 5, 4, 5]
Does 5 occur atleast 3 times ? : False
Time Complexity: O(n)
Auxiliary Space: O(1)
Method#7: Using Recursive method.
The algorithm is as follows:
- Check if the input list is empty or N is zero. If either of the conditions is true, return False.
- Check if the first element of the list is K. If it is, decrement N and recurse on the rest of the list.
- If the first element of the list is not K, recurse on the rest of the list with the same N.
- If N becomes zero at any point, return True.
- If the entire list has been processed and N is still not zero, return False.
Python3
def check_occurrences(lst, K, N):
if not lst or N == 0:
return False
if lst[0] == K:
return check_occurrences(lst[1:], K, N-1)
else:
return check_occurrences(lst[1:], K, N)
test_list = [1, 3, 5, 5, 4, 5]
print("The original list is : " + str(test_list))
k = 5
N = 3
res=check_occurrences(test_list, k, N)
print("Does %d occur atleast %d times ? : " % (k, N) + str(res))
|
Output
The original list is : [1, 3, 5, 5, 4, 5]
Does 5 occur atleast 3 times ? : False
The time complexity of the algorithm is O(n), where n is the length of the input list. In the worst case, the algorithm will have to check all the elements of the list.
The auxiliary space complexity of the algorithm is O(n), where n is the length of the input list. This is because in the worst case, the algorithm will have to recurse n times, which will result in n function calls on the call stack.
Method#8: Using reduce():
Algorithm:
- Define a function check_occurrences(lst, K, N) that takes a list lst, an integer K, and an integer N as input.
- Check the base case. If the list is empty or N is zero, return False.
- Otherwise, use the reduce function to count the number of occurrences of K in the list.
- Return True if the count is greater than or equal to N, otherwise return False.
- Initialize a list and the variables k and N.
- Call the function check_occurrences() with the list, k, and N.
- Print the result.
Python3
from functools import reduce
def check_occurrences(lst, K, N):
return reduce(lambda count, x: count + 1 if x == K else count, lst, 0) >= N
test_list = [1, 3, 5, 5, 4, 5]
print("The original list is : " + str(test_list))
k = 5
N = 3
res = check_occurrences(test_list, k, N)
print("Does %d occur at least %d times? : " % (k, N) + str(res))
|
Output
The original list is : [1, 3, 5, 5, 4, 5]
Does 5 occur at least 3 times? : True
Time Complexity: O(N), where N is the length of the input list.
Space Complexity: O(1), since we only use a constant amount of extra space.
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