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133 Discussions

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  • + 0 comments

    def pageCount(n, p): # Write your code here return min(p//2,(n//2)-(p//2))

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    one liner solution in python return int(min(p-0,(n+1 if n%2==0 else n)-p)/2)

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    #python solution
    def pageCount(n, p):
        total = math.floor(n/2)
        goal = math.floor(p/2)
        
        return min(goal,total-goal)
    
  • + 0 comments

    My solution in Python

    def pageCount(n, p):
        if p % 2 == 0 and n == p + 1:
            return 0
        
        if n % 2 == 0:
            n += 1
        
        return min(int(p/2), int((n - p)/2))
                
    
  • + 0 comments

    Python 3 solution:

    def pageCount(n: int, p: int) -> int:
        p -= p % 2
        return min(n - p, p) // 2
    
    
    pageCount=lambda n,p:min(p//2,n//2-p//2)  # 40