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Problem :

I am completely stuck on this problem

Given an array of ints, return True if the array contains a 2 next to a 2 somewhere.

has22([1, 2, 2])  True
has22([1, 2, 1, 2])  False
has22([2, 1, 2])  False

I know the basic idea but I am unable to implement it. I would also like to know that what type of the problem is this, eg. graph, search?

def has22(mynums):
for l in mynums:
    if ( (mynums[l] = 2) and (mynums[l+1] = 2) )
        return True
return False 
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7,540 points

1 Answer

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Solution :

Please find below the required source code:

def has22(mynums):
    for j in range(len(mynums) - 1):
        if mynums[j] == 2 and mynums[j+ 1] == 2:
            return True
    return False

This was the easiest solution that I came up with.

Using the for loop to check if the iterated number, mynums[j] == 2 "and" the one very next to it, which is [j+1] == 2 as well.

Here (len(mynums)-1): this line will just prevent it from going out of the range through the for loop as the j+1 on the final loop will check out of the range.

 

I hope you have clearly understood the solution given to you by me.

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38,600 points

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