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868. Binary Gap

Description

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

Solutions

Solution 1: Bit Manipulation

We use two pointers \(\textit{pre}\) and \(\textit{cur}\) to represent the positions of the previous and current \(1\) bits, respectively. Initially, \(\textit{pre} = 100\) and \(\textit{cur} = 0\). Then, we traverse the binary representation of \(n\). When we encounter a \(1\), we calculate the distance between the current position and the previous \(1\) position and update the answer.

The time complexity is \(O(\log n)\), where \(n\) is the given integer. The space complexity is \(O(1)\).

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class Solution:
    def binaryGap(self, n: int) -> int:
        ans = 0
        pre, cur = inf, 0
        while n:
            if n & 1:
                ans = max(ans, cur - pre)
                pre = cur
            cur += 1
            n >>= 1
        return ans

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class Solution {
    public int binaryGap(int n) {
        int ans = 0;
        for (int pre = 100, cur = 0; n != 0; n >>= 1) {
            if (n % 2 == 1) {
                ans = Math.max(ans, cur - pre);
                pre = cur;
            }
            ++cur;
        }
        return ans;
    }
}

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class Solution {
public:
    int binaryGap(int n) {
        int ans = 0;
        for (int pre = 100, cur = 0; n != 0; n >>= 1) {
            if (n & 1) {
                ans = max(ans, cur - pre);
                pre = cur;
            }
            ++cur;
        }
        return ans;
    }
};

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func binaryGap(n int) (ans int) {
    for pre, cur := 100, 0; n != 0; n >>= 1 {
        if n&1 == 1 {
            ans = max(ans, cur-pre)
            pre = cur
        }
        cur++
    }
    return
}

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function binaryGap(n: number): number {
    let ans = 0;
    for (let pre = 100, cur = 0; n; n >>= 1) {
        if (n & 1) {
            ans = Math.max(ans, cur - pre);
            pre = cur;
        }
        ++cur;
    }
    return ans;
}

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impl Solution {
    pub fn binary_gap(mut n: i32) -> i32 {
        let mut ans = 0;
        let mut pre = 100;
        let mut cur = 0;
        while n != 0 {
            if n % 2 == 1 {
                ans = ans.max(cur - pre);
                pre = cur;
            }
            cur += 1;
            n >>= 1;
        }
        ans
    }
}

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