Re: checking if triplets of integers occur in a list - why does my code give the wrong answer for one specific test case?

Here is the task I am given:

Write a function find_if_triplet(), which receives a single parameter, my_list, a list of integers, with values in ascending order.

The function returns a single boolean, has_triplet, which is set to True if an element of the list my_list appears at least three times. It returns False otherwise.

For instance,
if my_list = [1,1,1], the function should return has_triplet = True, if my_list = [1,2,3,3,3], the function should return has_triplet = True, if my_list = [1,2,2,2,2,3], the function should return has_triplet = True, if my_list = [1,2,2,3,4,4,5], the function should return has_triplet = False.

I tried to use the Python count() function to check if an integer in the list occurs at least thrice. While it works for most test cases i.e. when my_list = [1,1,1] or [1,2,3,3,3] or [1,2,2,3,4,4,5], when my_list =[1,2,2,2,2,3], it doesn't. Instead of the boolean returning a True (has triplets) it incorrectly returns False. Why could this be?

def find_if_triplet(my_list):
    for i in my_list:
    instances = my_list.count(i)
    if instances >= 3:
        has_triplet = True
    
    else:
        has_triplet = False

return has_triplet

The indentation in the original question is flawed. What's probably needed is this:

from collections import Counter

def find_if_triplet(my_list: list[int]) -> bool:
    if len(my_list) > 2:
        for x in set(my_list):
            if my_list.count(x) > 2:
                return True
    return False

# if you prefer Counter to do all the work for you then:

def find_if_triplet2(my_list: list[int]) -> bool:
    if len(my_list) > 2:
        _, n = Counter(my_list).most_common()[0]
        return n > 2
    return False
    

lst = [1,2,2,2,2,3]

print(find_if_triplet(lst))
print(find_if_triplet2(lst))

Output:

True
True

Note:

The fact that the input list is sorted (ascending) is irrelevant in this implementation

since you are looping through the list and then using list.count on each element, then for each element this function count the frequency of that number in entire array for each value.

also in your code, if for any value lets say in [1,2,2,2,2,4] for each value has_triplet value changes as (for each index) [False, True, True, True, True, False]
since you are taking the final value of has_triplet as return, you are getting final value ie wheter if last number is has_triplet True or not only.

so to solve this issue, you can use counter fujnction or make your counter function as

def find_if_triplet(array):
    has_triplet = False
    freq = {}
    for num in array:
        if num not in freq:
            freq[num] = 0
        freq[num] += 1
        if freq[num] >= 3:
            has_triplet = True
            break # once found triplet stop search further and end the looping
    return has_triplet