Return True if array contains a 2 or a 3

Question

I'm having trouble with this CodingBat problem: Given an int array length 2, return True if it contains a 2 or a 3. I've tried two different ways to solve this. Can anyone explain what I'm doing wrong? ``` #This one says index is out of range, why? def has23(nums): for i in nums: if nums[i]==2 or nums[i]==3: return True else: return False ``` ``` #This one doesn't past the test if a user entered 4,3. #It would yield False when it should be true. Why? def has23(nums): for i in nums: if i==2 or i==3: return True else: return False ```

Accepted Answer

Your first one doesn't work because the `for` loop in Python isn't the same as the `for` loop in other languages. Instead of iterating over the indices, it iterates over the actual elements.

`for item in nums` is roughly equivalent to:

```
for (int i = 0; i < nums.length; i++) {
    int item = nums[i];

...
}
```

Your second one doesn't work because it returns `False` too soon. If the loop encounters a value that isn't `2` or `3`, it returns `False` and doesn't loop through any other elements.

Change your loop to this:

```
def has23(nums):
    for i in nums:
        if i == 2 or i == 3:
            return True  # Only return `True` if the value is 2 or 3

return False  # The `for` loop ended, so there are no 2s or 3s in the list.
```

Or just use `in`:

```
def has23(nums):
    return 2 in nums or 3 in nums
```

Return True if array contains a 2 or a 3

Solution

Problem