Python dictionaries are really frail, and if you know how they work, you can easily construct input data that makes them take O(n^2) time. The reason there was a difference between your submissions is because your code is different between the two submission. In the one that survived, you set cnt[1]=0, which broke the hack in test case 27 (since it relies on key=1 not being in the dictionary until the very end).

Here is an anti-hash hack generator for Python dictionaries (it supports CPython dicts, and PyPy dicts and sets)

def anti_hash_hack(n, x=1, cpython=False):
    """
    Input: integer n > 0
    Output: List A of length n 
    such that x <= A[i] <= x + 2**(n.bit_length() + 2)
    """
    pow2 = 2**(n.bit_length() + 2)
    
    A = [x + pow2]
    i = perturb = x
    while len(A) < n//2:
        A.append(i + pow2)
        if cpython:
            perturb >>= 5
        i = (5 * i + perturb + 1) % pow2
        if not cpython:
            perturb >>= 5
    
    while len(A) < n:
        A.append(x)
    
    return A

To generate a hack for both of your submissions, I had to make $$$x \ge 3$$$ since you have cnt[1] = 0 and cnt[2] = 0 in your code.

n = 2*10**5
A = anti_hash_hack(n, 3, False)
print(1)
print(n)
print(*A)