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    C# return p <= n / 2 ? p / 2 : (n/2 -p/2);

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    C++ one liner:

    int pageCount(int n, int p) { return std::min(p/2, ((n/2)-(p/2))); }

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    My humble python solution:

    # Write your code here
    
    #left to right
    min_left = p // 2
    
    #right to left
    n = n + 1 if n % 2 == 0 else n
    min_right = (n - p) // 2
    
    return min_left if min_left <= min_right else min_right
    
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    Typescript Solution:

    const pageCount = (n: number, p: number): number => Math.min(Math.floor(p / 2), Math.floor(n / 2) - Math.floor(p / 2));
    
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    def pageCount(n, p):
        # Write your code here
        start = p // 2
        end = (n - p) // 2
        return start if start < end else end