LeetCode-Bitwise AND of Numbers Range

Posted 2020-08-08 LiBlog

Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.

For example, given the range [5, 7], you should return 4.

Credits:
Special thanks to @amrsaqr for adding this problem and creating all test cases.

 

 Analysis:

 

Find out the largest identical part starting from left of m and n, e.g.,

m = 20000, n=20218

1001110| 00100000
1001110| 10100000

Result:

1001110| 00000000

Solution:

public class Solution {
    public int rangeBitwiseAnd(int m, int n) {
        if (m>n || m==0) return 0;
        if (m==n) return m;
        
        int step =0;
        while (m!=n){
            m >>= 1;
            n >>= 1;
            step++;
        }
        return m<<=step;
        
    }
}

Solution 2:

If we do not know how to perform bit operator, we can still solve it like this:

public class Solution {
    public int rangeBitwiseAnd(int m, int n) {
        if (m>n || m==0) return 0;
        if (m==n) return m;
        
        StringBuilder mStr = new StringBuilder().append(Integer.toBinaryString(m));
        StringBuilder base = new StringBuilder();
        StringBuilder res = new StringBuilder();
        for (int i=0;i<mStr.length();i++){
            base.append(‘1‘);
            res.append(‘0‘);
        }
        
        for (int i=0;i<mStr.length();i++){
            if (mStr.charAt(i)==‘0‘){
                base.setCharAt(i,‘0‘);
            }
            if (mStr.charAt(i)==‘1‘){
                int baseVal = Integer.parseInt(base.toString(),2);
                if (n>baseVal){
                    return Integer.parseInt(res.toString(),2);
                } else {
                    res.setCharAt(i,‘1‘);
                }
            }
        }
        return Integer.parseInt(res.toString(),2);
        
    }
}