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:
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;
}
}