Welcome to Subscribe On Youtube

2376. Count Special Integers

Description

We call a positive integer special if all of its digits are distinct.

Given a positive integer n, return the number of special integers that belong to the interval [1, n].

 

Example 1:

Input: n = 20
Output: 19
Explanation: All the integers from 1 to 20, except 11, are special. Thus, there are 19 special integers.

Example 2:

Input: n = 5
Output: 5
Explanation: All the integers from 1 to 5 are special.

Example 3:

Input: n = 135
Output: 110
Explanation: There are 110 integers from 1 to 135 that are special.
Some of the integers that are not special are: 22, 114, and 131.

 

Constraints:

  • 1 <= n <= 2 * 109

Solutions

  • class Solution {
    public int countSpecialNumbers(int n) {
        List<Integer> digits = new ArrayList<>();
        while (n != 0) {
            digits.add(n % 10);
            n /= 10;
        }
        int m = digits.size();
        int ans = 0;
        for (int i = 1; i < m; ++i) {
            ans += 9 * A(9, i - 1);
        }
        boolean[] vis = new boolean[10];
        for (int i = m - 1; i >= 0; --i) {
            int v = digits.get(i);
            for (int j = i == m - 1 ? 1 : 0; j < v; ++j) {
                if (vis[j]) {
                    continue;
                }
                ans += A(10 - (m - i), i);
            }
            if (vis[v]) {
                break;
            }
            vis[v] = true;
            if (i == 0) {
                ++ans;
            }
        }
        return ans;
    }

    private int A(int m, int n) {
        return n == 0 ? 1 : A(m, n - 1) * (m - n + 1);
    }
}

  • class Solution {
public:
    int countSpecialNumbers(int n) {
        int ans = 0;
        vector<int> digits;
        while (n) {
            digits.push_back(n % 10);
            n /= 10;
        }
        int m = digits.size();
        vector<bool> vis(10);
        for (int i = 1; i < m; ++i) {
            ans += 9 * A(9, i - 1);
        }
        for (int i = m - 1; ~i; --i) {
            int v = digits[i];
            for (int j = i == m - 1 ? 1 : 0; j < v; ++j) {
                if (!vis[j]) {
                    ans += A(10 - (m - i), i);
                }
            }
            if (vis[v]) {
                break;
            }
            vis[v] = true;
            if (i == 0) {
                ++ans;
            }
        }
        return ans;
    }

    int A(int m, int n) {
        return n == 0 ? 1 : A(m, n - 1) * (m - n + 1);
    }
};

  • class Solution:
    def countSpecialNumbers(self, n: int) -> int:
        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


  • func countSpecialNumbers(n int) int {
	digits := []int{}
	for n != 0 {
		digits = append(digits, n%10)
		n /= 10
	}
	m := len(digits)
	vis := make([]bool, 10)
	ans := 0
	for i := 1; i < m; i++ {
		ans += 9 * A(9, i-1)
	}
	for i := m - 1; i >= 0; i-- {
		v := digits[i]
		j := 0
		if i == m-1 {
			j = 1
		}
		for ; j < v; j++ {
			if !vis[j] {
				ans += A(10-(m-i), i)
			}
		}
		if vis[v] {
			break
		}
		vis[v] = true
		if i == 0 {
			ans++
		}
	}
	return ans
}

func A(m, n int) int {
	if n == 0 {
		return 1
	}
	return A(m, n-1) * (m - n + 1)
}

All Problems

All Solutions