Finding Triangle numbers in the list

I need help to find numbers in list if it’s triangle number or not and if it’s triangle number, 1(string format), else 0(also string format).

For Example:

Input: n = [3, 4, 6, 55, 345]

Output: "10110"

Another Example:

Input: n = [0, 1, 2, 5]

Output: "1100"

Thanks in advance.

I am using Python 3.12.

Here’s my code:

n = [1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035]
a = int(input())
b = list(map(int,input().split()))
d = ""
for i in b:
    if i in n:
        d += "1"
    else:
        d += "0"
print(d)

But it’s limited by 1035. Is there any way to check it without using list? Maximum number is can be input is 10^8.

>Solution :

A triangular number or triangle number counts objects arranged in an equilateral triangle. Triangular numbers are a type of figurate number, other examples being square numbers and cube numbers. The nth triangular number is the number of dots in the triangular arrangement with n dots on each side, and is equal to the sum of the n natural numbers from 1 to n

https://en.wikipedia.org/wiki/Triangular_number

Top 20000 triangle numbers.. Similarly you can increase the limit depending upon your need.

As @Kelly suggested keep the elements in set

from itertools import accumulate
limit = 20000
a = set(accumulate(range(1, limit+1)))

For your question you can do:

n = [3, 4, 6, 55, 345]

''.join('1' if x in a else '0' for x in n )

#output
'10110'