LeetCode – Guess Number Higher or Lower (Java)

We are playing the Guess Game. The game is as follows:

I pick a number from 1 to n. You have to guess which number I picked.

Every time you guess wrong, I’ll tell you whether the number is higher or lower.

You call a pre-defined API guess(int num) which returns 3 possible results (-1, 1, or 0):

-1 : My number is lower
1 : My number is higher
0 : Congrats! You got it!
Example:
n = 10, I pick 6.

Return 6.

Java Solution

This is a typical binary search problem. Here is a Java solution.

public int guessNumber(int n) {
    int low=1;
    int high=n;
 
    while(low <= high){
        int mid = low+((high-low)/2);
        int result = guess(mid);
        if(result==0){
            return mid;
        }else if(result==1){
            low = mid+1;
        }else{
            high=mid-1;
        }
    }
 
    return -1;
}

What we learn from this problem?

low+(high-low)/2 yields the same value with (low+high)/2. However, the first expression is less expensive. In addition, the following expression can be used:

low+((high-low)>>1)
(low+high)>>>1

Under the assumption that high and low are both non-negative, we know for sure that the upper-most bit (the sign-bit) is zero.

So both high and low are in fact 31-bit integers.

high = 0100 0000 0000 0000 0000 0000 0000 0000 = 1073741824
low  = 0100 0000 0000 0000 0000 0000 0000 0000 = 1073741824

When you add them together they may “spill” over into the top-bit.

high + low =       1000 0000 0000 0000 0000 0000 0000 0000
           =  2147483648 as unsigned 32-bit integer
           = -2147483648 as signed   32-bit integer

(high + low) / 2   = 1100 0000 0000 0000 0000 0000 0000 0000 = -1073741824
(high + low) >>> 1 = 0100 0000 0000 0000 0000 0000 0000 0000 = 1073741824