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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0801_0900.s0868_binary_gap;

// #Easy #Bit_Manipulation #2022_05_16_Time_1_ms_(82.94%)_Space_41.4_MB_(28.50%)

/**
 * 868 - Binary Gap\.
 *
 * Easy
 *
 * Given a positive integer `n`, find and return _the **longest distance** between any two **adjacent**_ `1`_'s in the binary representation of_ `n`_. If there are no two adjacent_ `1`_'s, return_ `0`_._
 *
 * Two `1`'s are **adjacent** if there are only `0`'s separating them (possibly no `0`'s). The **distance** between two `1`'s is the absolute difference between their bit positions. For example, the two `1`'s in `"1001"` have a distance of 3.
 *
 * **Example 1:**
 *
 * **Input:** n = 22
 *
 * **Output:** 2
 *
 * **Explanation:** 22 in binary is "10110".
 *
 * The first adjacent pair of 1's is "10110" with a distance of 2.
 *
 * The second adjacent pair of 1's is "10110" with a distance of 1.
 *
 * The answer is the largest of these two distances, which is 2.
 *
 * Note that "10110" is not a valid pair since there is a 1 separating the two 1's underlined.
 *
 * **Example 2:**
 *
 * **Input:** n = 8
 *
 * **Output:** 0
 *
 * **Explanation:** 8 in binary is "1000". There are not any adjacent pairs of 1's in the binary representation of 8, so we return 0.
 *
 * **Example 3:**
 *
 * **Input:** n = 5
 *
 * **Output:** 2
 *
 * **Explanation:** 5 in binary is "101".
 *
 * **Constraints:**
 *
 * *   1 <= n <= 10⁹
**/
public class Solution {
    public int binaryGap(int n) {
        int max = 0;
        int pos = 0;
        int lastPos = -1;
        while (n != 0) {
            pos++;
            if ((n & 1) == 1) {
                if (lastPos != -1) {
                    max = Math.max(max, pos - lastPos);
                }
                lastPos = pos;
            }
            n >>= 1;
        }
        return max;
    }
}