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

Asked By: Jodi Peterson

||

Source

Answer 1

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

Answered By: Blender

Answer 2

def has23(nums):
  if nums[0] ==2 or nums[0]==3 or nums[1] ==2 or nums[1]==3: 
    return True
  else: 
    return False

Answered By: Saya

Answer 3

This Java code works fine:-

public boolean has23(int[] nums) {
  if(nums[0]==2||nums[1]==2||nums[0]==3||nums[1]==3){
    return true;
  }
  return false;
}

Answered By: ritesh_ratn

Answer 4

Another way to do the above using index
A variation for learning purpose

 def has23(nums):
  try :
    alpha = nums.index(2)
    return True
  except:
    try:
      beta = nums.index(3)
      return True
    except:
      return False

Answered By: srinivas chilukuri

Answer 5

Old post, I know, but for future readers:

Regarding the for loop, I think it’s worth mentioning another option: using the range() function.

Instead of

 for i in nums:

You can switch the for loop to look like so:

for i in range(len(nums)):

This will iterate over integers, the same as other languages do.
Then using nums[i] will get the value of the index.

However, I notice another problem with your code and its stated objectives:
within the for loop, all execution paths return a variable. It will only go through the for loop once, regardless of the length of the array, because after the first execution it returns, ending the function’s execution. If the first value is false, the function will return false.

Instead, you would want to end the execution within the loop only if the statement is true. If the loop goes through all posibilities and nothing is false, then return false:

def has23(nums):
 for i in range(len(nums)):   # iterate over the range of values
  if nums[i]==2 or nums[i]==3:# get values via index
   return true                # return true as soon as a true statement is found
 return false                 # if a true statement is never found, return false

Answered By: Astrolamb

Answer 6

def has23(nums):
  return (nums[0]==2 or nums[0]==3 or nums[1]==2 or nums[1]==3)

Answered By: lit31

Return True if array contains a 2 or a 3

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