Need Help in finding edge case

Правка en2, от SainathDora, 2025-05-10 19:02:50

Problem
Submission
What am I missing?

if(remaining_elements%2==0){
            cout<<(x+remaining_elements)<<"\n";
        }else{
            if(on_bits<=1){
                if(remaining_elements==1){
                    if(x==1){
                        cout<<(x+4)<<"\n";
                    }else{
                        cout<<(x+n)<<"\n";
                    }
                }else{
                    cout<<( x+remaining_elements+3)<<"\n";
                }
            }else{
                cout<<(x+remaining_elements+1)<<"\n";
            }
        }
    }


cout<<(x+4)<<"\n" this is pretty much handles a case, if x==1 and there are only one remaining_element i.e n=2

$$$x = 001$$$
$$$a = 011$$$
$$$b = 010$$$


cout<<(x+n)<<"\n", this is when $$$x=2^i$$$ where $$$i \gt 0$$$ and remaining_elements = 1 i.e when n=2

$$$x = 10000$$$
$$$a = 10001$$$
$$$b = 00001$$$


cout<<(x+remaining_elements+3)<<"\n", this handles when we have $$$x=2^i$$$ for $$$i \gt 0$$$ and remaining_elements = {3, 5, ...}

$$$x = 01000$$$
$$$a = 01000$$$
$$$b = 00001$$$
$$$c = 00011$$$
$$$d = 00010$$$


We can generalize this, since remaining_elements will be odd $$$ \gt =3$$$ we can set 0th bit on for all remaining_elements, upto
(remaining_elements-1) then we set 1st bit($$$2^1$$$) for last element and last second so in total : $$$x+(RemainingElements-1)+2+2$$$

$$$ans = x+RemainingElements+3$$$

so where does this fail?

Теги xor, xor properties, bitwise operators, bitwise

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  Rev. Язык Кто Когда Δ Комментарий
en2 Английский SainathDora 2025-05-10 19:02:50 923 Tiny change: '(remaining_elements-1)' -> '(remainingElements-1)' (published)
en1 Английский SainathDora 2025-05-10 08:35:34 1501 Initial revision (saved to drafts)