Returning int as None

Question

I’m returning count here, which before the return statement gives <type 'int'> but after it gets returned the output I get is None.

def equal(arr,count):
 arr = sorted(arr)
 if arr[0]==arr[-1]:
    
    print(count) # the answer I’m getting is correct so I don’t 
                  #think there’s a problem with rest of the code
    return count

 if (arr[-1]-arr[0])>=5:
     diff=5
 elif(arr[-1]-arr[0])>=2:
     diff=2
 else:
     diff=1
 for i in range(len(arr)-1):
     arr[i]+=diff

 count+=1
 equal(arr,count)


 a=[10,7,12]
 print(equal(a,0)) # I'm getting output here as **None**enter code here

Answer 1

Your function returns only if the if condition, is true as in lines :

 if arr[0]==arr[-1]:
    
    print(count) # the answer I’m getting is correct so I don’t 
                  #think there’s a problem with rest of the code
    return count

But if the condition, arr[0]==arr[-1] is false, your function does not return anything, that is why it is returning a None type. Add a return when the function calls itself in the body, with the line equal(arr,count) changed to return equal(arr,count)

def equal(arr,count):
    arr = sorted(arr)
    if arr[0]==arr[-1]:
    
        print(count) # the answer I’m getting is correct so I don’t 
                  #think there’s a problem with rest of the code
        return count

    if (arr[-1]-arr[0])>=5:
        diff=5
    elif(arr[-1]-arr[0])>=2:
        diff=2
    else:
        diff=1
    for i in range(len(arr)-1):
        arr[i]+=diff

    count+=1
    return equal(arr,count)


a=[10,7,12]
print(equal(a,0)) # I'm getting output here as **None**enter code here

Answer 2

Your if clause fails, and then you are not returning anything. In that case, python implicitly returns None. Add an else clause to return something in case the if fails