I am trying to solve a leetcode problem, in which I have to find the number in a list that does not repeat. I have written this code, but it fails in this test case. I cannot seem to understand why it fails in this case. Can someone give me an explanation?

this is the code:

class Solution:
    def singleNumber(self, nums: List[int]) -> int:

        i = 0
        while True:

            if nums.count(nums[i]) == 1:
                return nums[i]
                break
            else:
                nums.remove(nums[i])
                i = i   1

After passing 28 test cases it fails the case where nums = [1, 0, 1]

I have found another solution, so I am most interested in an explanation of why this code fails the given test case. Because I honestly don't understand it. Thanks.

CodePudding user response：

I'm surprised that this code passed 28 test cases without triggering the bug. The way that you use remove is buggy, and also not needed. I don't see any good reason to mutate the list.

The core problem is that you seem to think that remove removes all occurrences on an item. It doesn't -- it only removes the first occurrence.

With the test case that it fails, [1,0,1] you start with i = 0, hence you look at the first 1. Its count is > 1, so you call the remove method with it. The resulting list is [0,1]. At this pointi is incremented to 1, which is the index of 1 in the list. Since the first 1 has been removed, the count of the remaining 1 is 1. Your code thus returns 1 as the answer, rather than the correct answer of 0.

As a general rule of thumb, it is a bad idea to mutate a list while iterating over it. The way that indices of items can shift because of the list mutation is a frequent source of bugs.