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

Solution 1

Python Code
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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

Java Code
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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);
    }
}

C++ Code
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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);
    }
};

Go Code
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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)
}

Solution 2

Python Code
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class Solution:
    def countSpecialNumbers(self, n: int) -> int:
        return self.f(n)

    def f(self, n):
        @cache
        def dfs(pos, mask, lead, limit):
            if pos <= 0:
                return lead ^ 1
            up = a[pos] if limit else 9
            ans = 0
            for i in range(up + 1):
                if (mask >> i) & 1:
                    continue
                if i == 0 and lead:
                    ans += dfs(pos - 1, mask, lead, limit and i == up)
                else:
                    ans += dfs(pos - 1, mask | 1 << i, False, limit and i == up)
            return ans

        a = [0] * 11
        l = 0
        while n:
            l += 1
            a[l] = n % 10
            n //= 10
        return dfs(l, 0, True, True)

Java Code
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class Solution {
    private int[] a = new int[11];
    private int[][] dp = new int[11][1 << 11];

    public int countSpecialNumbers(int n) {
        return f(n);
    }

    private int f(int n) {
        for (var e : dp) {
            Arrays.fill(e, -1);
        }
        int len = 0;
        while (n > 0) {
            a[++len] = n % 10;
            n /= 10;
        }
        return dfs(len, 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] != -1) {
            return dp[pos][mask];
        }
        int up = limit ? a[pos] : 9;
        int ans = 0;
        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;
    }
}

C++ Code
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class Solution {
public:
    int a[11];
    int dp[11][1 << 11];

    int countSpecialNumbers(int n) {
        return f(n);
    }

    int f(int n) {
        memset(dp, -1, sizeof dp);
        int len = 0;
        while (n) {
            a[++len] = n % 10;
            n /= 10;
        }
        return dfs(len, 0, true, true);
    }

    int dfs(int pos, int mask, bool lead, bool limit) {
        if (pos <= 0) {
            return lead ? 0 : 1;
        }
        if (!lead && !limit && dp[pos][mask] != -1) {
            return dp[pos][mask];
        }
        int up = limit ? a[pos] : 9;
        int ans = 0;
        for (int i = 0; i <= up; ++i) {
            if ((mask >> i) & 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;
    }
};

Go Code
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func countSpecialNumbers(n int) int {
	return f(n)
}

func f(n int) int {
	a := make([]int, 11)
	dp := make([][]int, 11)
	for i := range dp {
		dp[i] = make([]int, 1<<11)
		for j := range dp[i] {
			dp[i][j] = -1
		}
	}
	l := 0
	for n > 0 {
		l++
		a[l] = n % 10
		n /= 10
	}
	var dfs func(int, int, bool, bool) int
	dfs = func(pos, mask int, lead, limit bool) int {
		if pos <= 0 {
			if lead {
				return 0
			}
			return 1
		}
		if !lead && !limit && dp[pos][mask] != -1 {
			return dp[pos][mask]
		}
		ans := 0
		up := 9
		if limit {
			up = a[pos]
		}
		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(l, 0, true, true)
}