868. Binary Gap

/**
 * Problem: LeetCode 868 - Binary Gap
 * Difficulty: Easy
 * Language: Java
 * Author: CodeByTushu
 */
class Solution {
    public int binaryGap(int n) {
        int lastPos = -1;
        int maxGap = 0;
        int currPos = 0;

        while (n > 0) {
            if ((n & 1) == 1) {
                if (lastPos != -1) {
                    maxGap = Math.max(maxGap, currPos - lastPos);
                }
                lastPos = currPos;
            }
            n >>= 1;
            currPos++;
        }

        return maxGap;
    }
}

"""
Problem: LeetCode 868 - Binary Gap
Difficulty: Easy
Language: Python
Author: CodeByTushu
"""
class Solution:
    def binaryGap(self, n: int) -> int:
        last_pos = -1
        max_gap = 0
        curr_pos = 0

        while n > 0:
            if n & 1:
                if last_pos != -1:
                    max_gap = max(max_gap, curr_pos - last_pos)
                last_pos = curr_pos
            n >>= 1
            curr_pos += 1

        return max_gap

/**
 * Problem: LeetCode 868 - Binary Gap
 * Difficulty: Easy
 * Language: C++
 * Author: CodeByTushu
 */
#include <algorithm>

using namespace std;

class Solution {
public:
    int binaryGap(int n) {
        int lastPos = -1;
        int maxGap = 0;
        int currPos = 0;

        while (n > 0) {
            if (n & 1) {
                if (lastPos != -1) {
                    maxGap = max(maxGap, currPos - lastPos);
                }
                lastPos = currPos;
            }
            n >>= 1;
            currPos++;
        }

        return maxGap;
    }
};

/**
 * Problem: LeetCode 868 - Binary Gap
 * Difficulty: Easy
 * Language: JavaScript
 * Author: CodeByTushu
 */
/**
 * @param {number} n
 * @return {number}
 */
var binaryGap = function(n) {
    let lastPos = -1;
    let maxGap = 0;
    let currPos = 0;

    while (n > 0) {
        if ((n & 1) === 1) {
            if (lastPos !== -1) {
                maxGap = Math.max(maxGap, currPos - lastPos);
            }
            lastPos = currPos;
        }
        n >>>= 1;
        currPos++;
    }

    return maxGap;
};