package com.fishercoder.solutions;

/**
 * 201. Bitwise AND of Numbers Range
 *
 * 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.
 */
public class _201 {

    public static class Solution1 {
        /**
         * credit: https://discuss.leetcode.com/topic/28538/java-python-easy-solution-with-explanation
         * Bitwise AND operation within range actually turns out to be doing some operations with just these two boundary numbers: m and n
         * e.g. 5 and 7, in binary, they are 101 and 111, the result for [5,7] is 5&6&7 which is 101&110&111
         * this we can understand it to be shifting the digits of m and n from left to right until they become the same, then we pad that number with zeroes on the right side
         */
        public int rangeBitwiseAnd(int m, int n) {
            int i = 0;
            while (m != n) {
                m >>= 1;
                n >>= 1;
                i++;
            }
            return (n << i);
        }
    }
}

package com.fishercoder.solutions;

/**
 * 201. Bitwise AND of Numbers Range
 *
 * Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.
 *
 * Example 1:
 * Input: [5,7]
 * Output: 4
 *
 * Example 2:
 * Input: [0,1]
 * Output: 0
 *
 */
public class _201 {

    public static class Solution1 {
        /**
         * credit: https://discuss.leetcode.com/topic/28538/java-python-easy-solution-with-explanation
         * Bitwise AND operation within range actually turns out to be doing some operations with just these two boundary numbers: m and n
         * e.g. 5 and 7, in binary, they are 101 and 111, the result for [5,7] is 5&6&7 which is 101&110&111
         * this we can understand it to be shifting the digits of m and n from left to right until they become the same, then we pad that number with zeroes on the right side
         */
        public int rangeBitwiseAnd(int m, int n) {
            int i = 0;
            while (m != n) {
                m >>= 1;
                n >>= 1;
                i++;
            }
            return (n << i);
        }
    }
}

refactor 201