Formatted question description: https://leetcode.ca/all/1012.html

1012. Complement of Base 10 Integer (Easy)

Every non-negative integer N has a binary representation.  For example, 5 can be represented as "101" in binary, 11 as "1011" in binary, and so on.  Note that except for N = 0, there are no leading zeroes in any binary representation.

The complement of a binary representation is the number in binary you get when changing every 1 to a 0 and 0 to a 1.  For example, the complement of "101" in binary is "010" in binary.

For a given number N in base-10, return the complement of it's binary representation as a base-10 integer.

 

Example 1:

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

Example 2:

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

Example 3:

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

 

Note:

  1. 0 <= N < 10^9

Companies:
Cloudera

Related Topics:
Math

Solution 1.

// OJ: https://leetcode.com/problems/complement-of-base-10-integer/

// Time: O(1)
// Space: O(1)
class Solution {
public:
    int bitwiseComplement(int N) {
        if (!N) return 1;
        unsigned mask = ~0;
        while (mask & N) mask <<= 1;
        return ~N & ~mask;
    }
};

Java

class Solution {
    public int numDupDigitsAtMostN(int N) {
        if (N <= 10)
            return 0;
        int duplicates = 0;
        int length = String.valueOf(N).length();
        for (int i = 2; i < length; i++)
            duplicates += dupNums(i);
        Set<Integer> usedDigits = new HashSet<Integer>();
        int curNum = (int) Math.pow(10, length - 1);
        int curLength = length - 1;
        int unit = curNum;
        boolean flag = false;
        while (unit > 1) {
            int lastDigit = N / unit % 10;
            int upperBound = N / unit * unit;
            while (curNum < upperBound) {
                int curDigit = curNum / unit % 10;
                int curCount = 0;
                if (!usedDigits.contains(curDigit)) {
                    curCount = 1;
                    int start = 9 - usedDigits.size();
                    for (int i = 0; i < curLength; i++)
                        curCount *= start - i;
                }
                duplicates += unit - curCount;
                curNum += unit;
            }
            if (usedDigits.add(lastDigit)) {
                curLength--;
                unit /= 10;
            } else {
                flag = true;
                break;
            }
        }
        if (flag)
            duplicates += N - curNum + 1;
        else {
            while (curNum <= N) {
                int lastDigit = curNum % 10;
                if (usedDigits.contains(lastDigit))
                    duplicates++;
                curNum++;
            }
        }
        return duplicates;
    }

    public int dupNums(int length) {
        int distincts = 9;
        for (int i = 1; i < length; i++)
            distincts *= 10 - i;
        return 9 * (int) Math.pow(10, length - 1) - distincts;
    }
}

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