Bitwise and (or &) of a range - GeeksforGeeks

Bitwise and (or &) of a range - GeeksforGeeks
Given two non-negative long integers, x and y given x <= y, the task is to find bit-wise and of all integers from x and y, i.e., we need to compute value of x & (x+1) & … & (y-1) & y.7

An efficient solution is to follow following steps.
1) Find position of Most Significant Bit (MSB) in both numbers.
2) If positions of MSB are different, then result is 0.
3) If positions are same. Let positions be msb_p.
……a) We add 2msb_p to result.
……b) We subtract 2msb_p from x and y,
……c) Repeat steps 1, 2 and 3 for new values of x and y.

// Find position of MSB in n. For example if n = 17,

// then position of MSB is 4. If n = 7, value of MSB

// is 3

int msbPos(ll n)

{

    int msb_p = -1;

    while (n)

    {

        n = n>>1;

        msb_p++;

    }

    return msb_p;

}

 

// Function to find Bit-wise & of all numbers from x

// to y.

ll andOperator(ll x, ll y)

{

    ll res = 0; // Initialize result

 

    while (x && y)

    {

        // Find positions of MSB in x and y

        int msb_p1 = msbPos(x);

        int msb_p2 = msbPos(y);

 

        // If positions are not same, return

        if (msb_p1 != msb_p2)

            break;

 

        // Add 2^msb_p1 to result

        ll msb_val =  (1 << msb_p1);

        res = res + msb_val;

 

        // subtract 2^msb_p1 from x and y.

        x = x - msb_val;

        y = y - msb_val;

    }

 

    return res;

}

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