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3950. Exactly One Consecutive Set Bits Pair

Description

You are given an integer n.

Return true if its binary representation contains exactly one pair of consecutive set bits, and false otherwise.

Β 

Example 1:

Input: nums = 6

Output: true

Explanation:

  • Binary representation of 6 is 110.
  • There is exactly one pair of consecutive set bits ("11"). Thus, the answer is true​​​​​​​.

Example 2:

Input: nums = 5

Output: false

Explanation:

  • Binary representation of 5 is 101.
  • There are no consecutive set bits. Thus, the answer is false​​​​​​​.

Β 

Constraints:

  • 0 <= n <= 105

Solutions

Solution 1: Simulation

We use a variable \(\textit{pre}\) to record the digit of the previous bit, initialized to \(\textit{pre} = 0\), and another variable \(\textit{vis}\) to record whether a pair of consecutive set bits has already been found, initialized to \(\textit{vis} = \text{false}\).

Iterate through each binary bit of \(n\), and denote the current binary bit as \(\textit{cur}\). If \(\textit{pre} = \textit{cur} = 1\), and if \(\textit{vis} = \text{true}\) at this moment, it indicates that there are multiple pairs of consecutive set bits, so we directly return \(\text{false}\). Otherwise, we set \(\textit{vis}\) to \(\text{true}\). Then, we update \(\textit{pre} = \textit{cur}\) and continue to iterate through the next binary bit.

After the iteration ends, if \(\textit{vis} = \text{true}\), return \(\text{true}\); otherwise, return \(\text{false}\).

The time complexity is \(O(\log n)\), and the space complexity is \(O(1)\).

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class Solution:
    def consecutiveSetBits(self, n: int) -> bool:
        pre = 0
        vis = False
        while n:
            cur = n & 1
            if pre == cur == 1:
                if vis:
                    return False
                vis = True
            pre = cur
            n = n >> 1
        return vis

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class Solution {
    public boolean consecutiveSetBits(int n) {
        boolean vis = false;
        for (int pre = 0; n > 0; n >>= 1) {
            int cur = n & 1;
            if (pre == cur && cur == 1) {
                if (vis) {
                    return false;
                }
                vis = true;
            }
            pre = cur;
        }
        return vis;
    }
}

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class Solution {
public:
    bool consecutiveSetBits(int n) {
        bool vis = false;
        for (int pre = 0; n > 0; n >>= 1) {
            int cur = n & 1;
            if (pre == cur && cur == 1) {
                if (vis) {
                    return false;
                }
                vis = true;
            }
            pre = cur;
        }
        return vis;
    }
};

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func consecutiveSetBits(n int) bool {
    vis := false
    for pre := 0; n > 0; n >>= 1 {
        cur := n & 1
        if pre == cur && cur == 1 {
            if vis {
                return false
            }
            vis = true
        }
        pre = cur
    }
    return vis
}

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function consecutiveSetBits(n: number): boolean {
    let vis = false;
    for (let pre = 0; n > 0; n >>= 1) {
        const cur = n & 1;
        if (pre === cur && cur === 1) {
            if (vis) {
                return false;
            }
            vis = true;
        }
        pre = cur;
    }
    return vis;
}

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