Two red beads are placed between every two blue beads. There are N blue beads. After looking at the arrangement below work out the number of red beads.

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Implement count_red_beads(n) (in PHP count_red_beads($n); in Java, Javascript, TypeScript, C, C++ countRedBeads(n)) so that it returns the number of red beads.
If there are less than 2 blue beads return 0.

Difficulty: introductory

Starter code:

def count_red_beads(n):
	

Solutions:
def count_red_beads(nb):
    return max(0, 2 * (nb - 1) )

def count_red_beads(n_blue):
    return (n_blue - 1) * 2 if n_blue >= 2 else 0


def count_red_beads(n):
    return max(0, 2 * (n-1))
    #maximum value -> two for every two beads times number of blue beads - 1


def count_red_beads(n):
  return 2*(n - 1) if n > 1 else 0

count_red_beads=lambda b:b and b*2-2

def count_red_beads(n):
    return 2*max(n-1,0)

def count_red_beads(n):
    if n >= 2:
        red = n+(n-2)
        return red
    else: 
        return 0

def count_red_beads(N_blue):
    return (N_blue-2)*2+2 if N_blue >2 else 0

def count_red_beads(N_blue):
    return max(N_blue - 1, 0) * 2

def count_red_beads(n_blue):
    return 2 * (n_blue - 1) if n_blue > 1 else 0

def count_red_beads(n):
    return max(0, (n-1) * 2)

def count_red_beads(n):
    return (n-1)*2 if n > 1 else 0

def count_red_beads(n):
    return (n-1) * 2 if n else 0

def count_red_beads(N_blue):
    return 2*(N_blue - 1) if N_blue >= 2 else 0

def count_red_beads(N_b):
    return 0 if N_b == 0 else (N_b - 1) * 2

def count_red_beads(N_blue):
    return (N_blue - 1) * 2 if N_blue >= 2 else 0

count_red_beads = lambda n: n and (n - 1) * 2

def count_red_beads(N_blue):
    return N_blue and (N_blue-1)*2

def count_red_beads(n):
    return 0 if n == 0 else n*2 - 2

def count_red_beads(n):
    return n and n*2-2

count_red_beads=lambda b:(b>1)*(b-1)*2

count_red_beads=lambda n:max(n-1,0)*2

def count_red_beads(n):
    if n < 2:
        return 0
    elif n == 2:
        return 2
    else:
        return ((n-2) * 2) +2

def count_red_beads(n):
    return (n-1)*(n>0)*2

import math

def count_red_beads(n):

    if n<2:
    
        return 0
        
    else:
        
        if n%2!=0:
    
            g = math.floor(n/2)
        
            d = g*4
    
            return d
    
        if n%2==0:
    
            g = math.floor(n/2)
        
            d = g*4
    
            return d-2


def count_red_beads(n):
    if (n != 0):
        return 2 * (n - 1)
    return 0


def count_red_beads(n):
    lista = []
    if n < 2:
        return 0
    for i in range(1,n):
        lista.append(i*2)
    return lista[-1]
       


def count_red_beads(n):
    return 2*n-2*(n>0)

def count_red_beads(n):
    return (max(1,n)-1)*2 

def count_red_beads(n):
    return (n - 1) * 2 if n else n

def count_red_beads(n):
    if n<=1:
        return 0
    return 2*(n-1)

def count_red_beads(n):
    if n < 2:
        return 0
    if n == 2:
        return 2
    else:
        return n*2 - 2

def count_red_beads(n):
    if n == 0:
        return 0
    else:
        output = (n - 1) * 2
        
        return output

def count_red_beads(n):
    if n == 0 or n == 1:
        return 0
    else:
        return (n - 1) * 2

count_red_beads = lambda n: max(0, 2 * (n - 1))

def count_red_beads(n):
    return 0 if n < 2 else 2 * ((3*n-1)//3) 

def count_red_beads(n):
    return [0, (n-1)*2][n > 2]

def count_red_beads(n):
    if n==1 or n==0:
        return 0
    return (n-1)*2

count_red_beads = lambda n: 0 if n in [0,1] else (n-1)*2

def count_red_beads(n):
    return range(0,10001,2)[n-1 if n>0 else n]

def count_red_beads(n):
    if n-2<0:
        return 0
    else:
        y= n*2
        z= y-2
        return(z)


def count_red_beads(n):
    return max(0, n - 1 << 1)

def count_red_beads(n):
    if n-3<0:
        return 0
    else:
        y= n*2
        z= y-2
        return(z)

def count_red_beads(n):
    counter = 0
    if n < 2:
        return counter
    else:
        for x in range(1, n):
            counter +=2
        return counter

def count_red_beads(n):
    if n>=2:
        return (2*n)-2
    return 0

def count_red_beads(n):
    if n <= 0:
        return 0

    return (n - 1) * 2

def count_red_beads(n):
    if n < 2:
        return 0
    red = (n - 1) * 2
    return red

count_red_beads = lambda n: (n - 1) * 2 if n else 0

def count_red_beads(n):
    if n < 2:
        return 0
    elif n % 2 == 0:
        return (n - 1) * 2
    else:
        return (n * 2) - 2

def count_red_beads(n):
    if n < 2:
        r = 0
    else:
        r = (n*2) - 2
    return(r)

def count_red_beads(n):
    if n <= 1:
        return 0
    elif n > 1:
        return (n*2)-2


def count_red_beads(n):
    red_beads = 0
    if n < 2:
        return 0
    red_beads = (n-1) * 2
    return red_beads
        


def count_red_beads(n):
    return  0 if n-1 < 2 else(n - 1) * 2

def count_red_beads(n):
    return 0 if n < 2 else 2 if n == 2 else (n-1)*2

def count_red_beads(n):
    return (n - 1) * 2 if n > 1 else 0
    # Flez


def count_red_beads(blue):
    if blue <= 1:
        return 0
    return (blue - 1) * 2

def count_red_beads(n):
    if n < 1:
        return 0
    else:
        return n + (n-2)

def count_red_beads(n):
    if n < 2:
        return 0
    else:
        count = 2
        for i in range(2, n):
            count += 2

        return count

def count_red_beads(n):
    while n > 0:
        return (n *2) - 2
    return 0

def count_red_beads(n):
    if n >= 2:
        x = n -2
        result = n + x
        return result
    else:
        return 0

def count_red_beads(n):
  count = 0
  if n < 2:
    return 0
  for x in range(1,n):
    count += 2
  return count

def count_red_beads(n):
    return 'rr'.join(['b' for b in range(n)]).count('r')

def count_red_beads(n):
    return (max(n,1)-1) * 2

def count_red_beads(n):
    res=0
    for i in range(1,n):
        res+=2
    return res

def count_red_beads(n):
    if n < 2:
        return 0
    else:
        y = (n * 2) - 2
        return y

def count_red_beads(n):
    if n > 2:
        return 2*n-2
    else:
        return 0

def count_red_beads(n):
    if n < 2:
        return 0
        
    r = n * 2 - 2
    return r

def count_red_beads(n):
    if n < 2:
        return 0
    else:
        n -= 1
        output = n * 2
        return output

def count_red_beads(n):
    return (n-1)*2 if n>1 else 0  #I solved this Kata on  [28-Nov-2019] ^_^ [11:33 PM]...#Hussam'sCodingDiary

def count_red_beads(n):
    numOfRed = 0
    if n < 2:
        return 0
    else:
       numOfRed += n * 2 - 2
    return numOfRed

def count_red_beads(n):
    if n < 2:
        return 0
    elif n >= 2:
        return 2*(n-1)

def count_red_beads(n):
    r = 2
    if n < 2:
        return 0
    for x in range(2, n):
        r += 2
    return r

def count_red_beads(n):
    if (n - 1) * 2 < 0:
        return 0
    return (n - 1) * 2

count_red_beads = lambda n: max(0, (n-1)*2)

def count_red_beads(n):
    count = 0
    if n < 2:
        return count
    for i in range(2, n+1):
        count += 2
    return count

count_red_beads = lambda n: (n-1)*2*(n>0)

def count_red_beads(n):
    if n == 0:
         return 0
    else:
        return ((n*2)-2)

def count_red_beads(n):
  if n < 2:
    return 0
  else: 
    p = (n*2)-2
    return p

def count_red_beads(x):
    if x>1:
        return(2*(x-1))
    else:
        return(0)

def count_red_beads(n):
    if n == 2:
        return 0
    elif n>0:
        return (n-1)*2
    return 0


def count_red_beads(n):
    return [0, (n-1)*2][n>0]

def count_red_beads(n):
    return (n + n) - 2 if n > 0 else 0

count_red_beads = lambda n: (n - (1 if n else 0)) * 2

count_red_beads = lambda n: (0, (n - 1) * 2)[bool(n)]

def count_red_beads(n):
    if n > 1:
        red = 2*n - 2
        return red
    else:
        return 0

def count_red_beads(n):
    if n < 2:
        return 0
    else:
        red_beds = n * 2 - 2
    return red_beds

def count_red_beads(n):
    return n and (n-1)*2

def count_red_beads(n):
    if n < 2:
        return 0
    else:
        n -= 1
        return 2 * n

def count_red_beads(n):
    if n == 0 or n == 1:
        return 0
    return (n-1)*2

def count_red_beads(n):
    if n > 1:
        return n * 2 - 2
    elif n == 1:
        return 0
    elif n < 1:
        return 0

def count_red_beads(n):
    if n==1 or n==0:
        return 0
    elif n%2!=0:
        return (n-1)*2
    else:
        return(n*2)-2

def count_red_beads(n):
    a = -2
    for i in range(n):
        a += 2
    return (a if a > 0 else 0)

def count_red_beads(n):
    if n < 2:
        return 0
    if n > 2:
        red = (n-1)*2
        return red

def count_red_beads(n):
    if n<2:
        return 0
    else:
        i =0
        RedCount = -2
    while(i<n):
        RedCount+=2
        i+=1
    return RedCount


def count_red_beads(n):
    beads = 0
    if n < 2:
        return 0
    else: 
        for i in range(n-1):
            beads += 2 
    return beads 

def count_red_beads(n):
    ecount = 2
    ocount = 4
    oplst1 = []
    
    if n < 2:
        return 0
    else:
        for i in range(2, n+1):
            if i % 2 == 0:
                oplst1.append(ecount)
                ecount += 4
            else:
                oplst1.append(ocount)
                ocount += 4
    
        return oplst1[-1]

def count_red_beads(n):
    if n == 2:
        return n
    elif n > 2:
        return n * 2 - 2
    else:
        return 0

def count_red_beads(n):
    beads = ''
    for i in range(n-1):
        beads += 'B'
        beads += 'RR'
    beads += 'B'
#     print(beads)
    return beads.count('R')

def count_red_beads(n):
    return ((n or 1)-1)*2