def pageCount(n, p): return p//2 if p//2 <= n //4 else n//2 - p//2

def pageCount(n, p): if(n%2==0 and p==n-1 and p!=1): return 1 if((p//2)+1>(n-p)//2): return (n-p)//2 else: return p//2

In C++ Solution

int pageCount(int n, int p) { cin>>n>>p;

return min(p/2,n/2-p/2);

}

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Here is my c++ solution, you can watch the explanation here : https://youtu.be/1T650YeAwRI

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