357. Count Numbers with Unique Digits

Posted 2021-02-08 ruruozhenhao

Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.

Example:

Input: 2
Output: 91 
Explanation: The answer should be the total numbers in the range of 0 ≤ x < 100, 
             excluding 11, 22, 33, 44, 55, 66, 77, 88, 99

 

Approach #1: Math. [C++]

class Solution {
public:
    int countNumbersWithUniqueDigits(int n) {
        if (n == 0) return 1;
        if (n == 1) return 10;
        
        int ans = 10;
        int count = 9;
        int temp = 9;
        while (n > 1 && count > 0) {
            temp = temp * count;
            ans += temp;
            count--;
            n--;
        }
        
        return ans;
    }
};

　　

Analysis:

f(1) = 10;

f(2) = f(1) + 9 * 9;　　xy: x: can be 1, 2.....8, 9.  y: can‘t be the same as with x but it can be 0, so there are 9 difference kinds choise.

f(2) = f(2) + 9 * 9 * 8;

 

Approach #2: backtracking. [C++]

public class Solution {
	public static int countNumbersWithUniqueDigits(int n) {
		if (n > 10) {
			return countNumbersWithUniqueDigits(10);
		}
		int count = 1; // x == 0
		long max = (long) Math.pow(10, n);

		boolean[] used = new boolean[10];

		for (int i = 1; i < 10; i++) {
			used[i] = true;
			count += search(i, max, used);
			used[i] = false;
		}

		return count;
	}

	private static int search(long prev, long max, boolean[] used) {
		int count = 0;
		if (prev < max) {
			count += 1;
		} else {
			return count;
		}

		for (int i = 0; i < 10; i++) {
			if (!used[i]) {
				used[i] = true;
				long cur = 10 * prev + i;
				count += search(cur, max, used);
				used[i] = false;
			}
		}

		return count;
	}
}