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1009. Complement of Base 10 Integer

Description

The complement of an integer is the integer you get when you flip all the 0's to 1's and all the 1's to 0's in its binary representation.

  • For example, The integer 5 is "101" in binary and its complement is "010" which is the integer 2.

Given an integer n, return its complement.

 

Example 1:

Input: n = 5
Output: 2
Explanation: 5 is "101" in binary, with complement "010" in binary, which is 2 in base-10.

Example 2:

Input: n = 7
Output: 0
Explanation: 7 is "111" in binary, with complement "000" in binary, which is 0 in base-10.

Example 3:

Input: n = 10
Output: 5
Explanation: 10 is "1010" in binary, with complement "0101" in binary, which is 5 in base-10.

 

Constraints:

  • 0 <= n < 109

 

Note: This question is the same as 476: https://leetcode.com/problems/number-complement/

Solutions

Solution 1: Bit Manipulation

Thinking

Writing the binary form and flipping bits yields the complement, but \(n\) can be \(10^9\), leading zeros must not be flipped, and \(n=0\) is defined to be \(1\).

Processing from low to high only inverts bits that actually appear in \(n\). When \(n\) becomes \(0\) we stop, so higher zero bits stay untouched.

Index \(i\) marks the current bit; we OR the flipped low bit of \(n\) into \(\textit{ans}\) and shift \(n\) until it vanishes.

First, we check if \(n\) is \(0\). If it is, we return \(1\).

Next, we define two variables \(\textit{ans}\) and \(i\), both initialized to \(0\). Then we iterate through \(n\). In each iteration, we set the \(i\)-th bit of \(\textit{ans}\) to the inverse of the \(i\)-th bit of \(n\), increment \(i\) by \(1\), and right shift \(n\) by \(1\).

Finally, we return \(\textit{ans}\).

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

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class Solution:
    def bitwiseComplement(self, n: int) -> int:
        if n == 0:
            return 1
        ans = i = 0
        while n:
            ans |= ((n & 1 ^ 1) << i)
            i += 1
            n >>= 1
        return ans

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class Solution {
    public int bitwiseComplement(int n) {
        if (n == 0) {
            return 1;
        }
        int ans = 0, i = 0;
        while (n != 0) {
            ans |= (n & 1 ^ 1) << (i++);
            n >>= 1;
        }
        return ans;
    }
}

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class Solution {
public:
    int bitwiseComplement(int n) {
        if (n == 0) {
            return 1;
        }
        int ans = 0, i = 0;
        while (n != 0) {
            ans |= (n & 1 ^ 1) << (i++);
            n >>= 1;
        }
        return ans;
    }
};

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func bitwiseComplement(n int) (ans int) {
    if n == 0 {
        return 1
    }
    for i := 0; n != 0; n >>= 1 {
        ans |= (n&1 ^ 1) << i
        i++
    }
    return
}

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function bitwiseComplement(n: number): number {
    if (n === 0) {
        return 1;
    }
    let ans = 0;
    for (let i = 0; n; n >>= 1) {
        ans |= ((n & 1) ^ 1) << i++;
    }
    return ans;
}

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impl Solution {
    pub fn bitwise_complement(mut n: i32) -> i32 {
        if n == 0 {
            return 1;
        }
        let mut ans = 0;
        let mut i = 0;
        while n != 0 {
            ans |= ((n & 1) ^ 1) << i;
            n >>= 1;
            i += 1;
        }
        ans
    }
}

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public class Solution {
    public int BitwiseComplement(int n) {
        if (n == 0) {
            return 1;
        }
        int ans = 0, i = 0;
        while (n != 0) {
            ans |= ((n & 1) ^ 1) << (i++);
            n >>= 1;
        }
        return ans;
    }
}

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