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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.

  • 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;
        }
    }
    
    ############
    
    class Solution {
        private int[] nums = new int[11];
        private Integer[][] dp = new Integer[11][1 << 11];
    
        public int numDupDigitsAtMostN(int n) {
            return n - f(n);
        }
    
        private int f(int n) {
            int i = -1;
            for (; n > 0; n /= 10) {
                nums[++i] = n % 10;
            }
            return dfs(i, 0, true, true);
        }
    
        private int dfs(int pos, int mask, boolean lead, boolean limit) {
            if (pos < 0) {
                return lead ? 0 : 1;
            }
            if (!lead && !limit && dp[pos][mask] != null) {
                return dp[pos][mask];
            }
            int ans = 0;
            int up = limit ? nums[pos] : 9;
            for (int i = 0; i <= up; ++i) {
                if ((mask >> i & 1) == 1) {
                    continue;
                }
                if (i == 0 && lead) {
                    ans += dfs(pos - 1, mask, lead, limit && i == up);
                } else {
                    ans += dfs(pos - 1, mask | 1 << i, false, limit && i == up);
                }
            }
            if (!lead && !limit) {
                dp[pos][mask] = ans;
            }
            return ans;
        }
    }
    
  • // 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;
        }
    };
    
  • class Solution:
        def numDupDigitsAtMostN(self, n: int) -> int:
            return n - self.f(n)
    
        def f(self, n):
            def A(m, n):
                return 1 if n == 0 else A(m, n - 1) * (m - n + 1)
    
            vis = [False] * 10
            ans = 0
            digits = [int(c) for c in str(n)[::-1]]
            m = len(digits)
            for i in range(1, m):
                ans += 9 * A(9, i - 1)
            for i in range(m - 1, -1, -1):
                v = digits[i]
                j = 1 if i == m - 1 else 0
                while j < v:
                    if not vis[j]:
                        ans += A(10 - (m - i), i)
                    j += 1
                if vis[v]:
                    break
                vis[v] = True
                if i == 0:
                    ans += 1
            return ans
    
    ############
    
    class Solution(object):
        def bitwiseComplement(self, N):
            """
            :type N: int
            :rtype: int
            """
            return int("".join(map(lambda x :"0" if x == "1" else "1", bin(N)[2:])), 2)
    
  • func numDupDigitsAtMostN(n int) int {
    	return n - f(n)
    }
    
    func f(n int) int {
    	nums := []int{}
    	for ; n > 0; n /= 10 {
    		nums = append(nums, n%10)
    	}
    	dp := [11][1 << 11]int{}
    	for i := range dp {
    		for j := range dp[i] {
    			dp[i][j] = -1
    		}
    	}
    	var dfs func(int, int, bool, bool) int
    	dfs = func(pos int, mask int, lead bool, limit bool) int {
    		if pos < 0 {
    			if lead {
    				return 0
    			}
    			return 1
    		}
    		if !lead && !limit && dp[pos][mask] != -1 {
    			return dp[pos][mask]
    		}
    		up := 9
    		if limit {
    			up = nums[pos]
    		}
    		ans := 0
    		for i := 0; i <= up; i++ {
    			if mask>>i&1 == 1 {
    				continue
    			}
    			if i == 0 && lead {
    				ans += dfs(pos-1, mask, lead, limit && i == up)
    			} else {
    				ans += dfs(pos-1, mask|1<<i, false, limit && i == up)
    			}
    		}
    		if !lead && !limit {
    			dp[pos][mask] = ans
    		}
    		return ans
    	}
    	return dfs(len(nums)-1, 0, true, true)
    }
    
  • function numDupDigitsAtMostN(n: number): number {
        return n - f(n);
    }
    
    function f(n: number): number {
        const nums: number[] = [];
        let i = -1;
        for (; n; n = Math.floor(n / 10)) {
            nums[++i] = n % 10;
        }
        const dp = Array.from({ length: 11 }, () => Array(1 << 11).fill(-1));
        const dfs = (
            pos: number,
            mask: number,
            lead: boolean,
            limit: boolean,
        ): number => {
            if (pos < 0) {
                return lead ? 0 : 1;
            }
            if (!lead && !limit && dp[pos][mask] !== -1) {
                return dp[pos][mask];
            }
            const up = limit ? nums[pos] : 9;
            let ans = 0;
            for (let i = 0; i <= up; ++i) {
                if ((mask >> i) & 1) {
                    continue;
                }
                if (lead && i === 0) {
                    ans += dfs(pos - 1, mask, lead, limit && i === up);
                } else {
                    ans += dfs(pos - 1, mask | (1 << i), false, limit && i === up);
                }
            }
            if (!lead && !limit) {
                dp[pos][mask] = ans;
            }
            return ans;
        };
        return dfs(i, 0, true, true);
    }
    
    

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