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1866. Number of Ways to Rearrange Sticks With K Sticks Visible

SourceWeekly Contest 241 Q4DifficultyHardRating2333

Description

There are n uniquely-sized sticks whose lengths are integers from 1 to n. You want to arrange the sticks such that exactly k sticks are visible from the left. A stick is visible from the left if there are no longer sticks to the left of it.

  • For example, if the sticks are arranged [1,3,2,5,4], then the sticks with lengths 1, 3, and 5 are visible from the left.

Given n and k, return the number of such arrangements. Since the answer may be large, return it modulo 109 + 7.

 

Example 1:

Input: n = 3, k = 2
Output: 3
Explanation: [1,3,2], [2,3,1], and [2,1,3] are the only arrangements such that exactly 2 sticks are visible.
The visible sticks are underlined.

Example 2:

Input: n = 5, k = 5
Output: 1
Explanation: [1,2,3,4,5] is the only arrangement such that all 5 sticks are visible.
The visible sticks are underlined.

Example 3:

Input: n = 20, k = 11
Output: 647427950
Explanation: There are 647427950 (mod 109 + 7) ways to rearrange the sticks such that exactly 11 sticks are visible.

 

Constraints:

  • 1 <= n <= 1000
  • 1 <= k <= n

Solutions

Solution 1: Dynamic Programming

Thinking

Count permutations of \(n\) sticks where exactly \(k\) are visible from the left. A stick is visible iff it is the tallest so far. \(n\le 1000\) forbids listing permutations.

Let \(f[i][j]\) be the ways among \(i\) sticks with \(j\) visible. Placing the tallest last always adds a visible stick; otherwise the last height is one of the other \(i-1\) values and the visible count stays the same. The recurrence yields \(f[n][k]\).

We define \(f[i][j]\) to represent the number of permutations of length \(i\) in which exactly \(j\) sticks can be seen. Initially, \(f[0][0]=1\) and the rest \(f[i][j]=0\). The answer is \(f[n][k]\).

Consider whether the last stick can be seen. If it can be seen, it must be the longest. Then there are \(i - 1\) sticks in front of it, and exactly \(j - 1\) sticks can be seen, which is \(f[i - 1][j - 1]\). If the last stick cannot be seen, it can be any one except the longest stick. Then there are \(i - 1\) sticks in front of it, and exactly \(j\) sticks can be seen, which is \(f[i - 1][j] \times (i - 1)\).

Therefore, the state transition equation is:

\[ f[i][j] = f[i - 1][j - 1] + f[i - 1][j] \times (i - 1) \]

The final answer is \(f[n][k]\).

The time complexity is \(O(nk)\), and the space complexity is \(O(nk)\), where \(n\) and \(k\) are the two integers given in the problem.

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class Solution:
    def rearrangeSticks(self, n: int, k: int) -> int:
        mod = 10**9 + 7
        f = [[0] * (k + 1) for _ in range(n + 1)]
        f[0][0] = 1
        for i in range(1, n + 1):
            for j in range(1, k + 1):
                f[i][j] = (f[i - 1][j - 1] + f[i - 1][j] * (i - 1)) % mod
        return f[n][k]

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class Solution {
    public int rearrangeSticks(int n, int k) {
        final int mod = (int) 1e9 + 7;
        int[][] f = new int[n + 1][k + 1];
        f[0][0] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= k; ++j) {
                f[i][j] = (int) ((f[i - 1][j - 1] + f[i - 1][j] * (long) (i - 1)) % mod);
            }
        }
        return f[n][k];
    }
}

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class Solution {
public:
    int rearrangeSticks(int n, int k) {
        const int mod = 1e9 + 7;
        int f[n + 1][k + 1];
        memset(f, 0, sizeof(f));
        f[0][0] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= k; ++j) {
                f[i][j] = (f[i - 1][j - 1] + (i - 1LL) * f[i - 1][j]) % mod;
            }
        }
        return f[n][k];
    }
};

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func rearrangeSticks(n int, k int) int {
    const mod = 1e9 + 7
    f := make([][]int, n+1)
    for i := range f {
        f[i] = make([]int, k+1)
    }
    f[0][0] = 1
    for i := 1; i <= n; i++ {
        for j := 1; j <= k; j++ {
            f[i][j] = (f[i-1][j-1] + (i-1)*f[i-1][j]) % mod
        }
    }
    return f[n][k]
}

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function rearrangeSticks(n: number, k: number): number {
    const mod = 10 ** 9 + 7;
    const f: number[][] = Array.from({ length: n + 1 }, () =>
        Array.from({ length: k + 1 }, () => 0),
    );
    f[0][0] = 1;
    for (let i = 1; i <= n; ++i) {
        for (let j = 1; j <= k; ++j) {
            f[i][j] = (f[i - 1][j - 1] + (i - 1) * f[i - 1][j]) % mod;
        }
    }
    return f[n][k];
}

Solution 2: Dynamic Programming (Space Optimization)

Thinking

Solution 1's \(f[i][j]\) depends only on the previous row. Compress to one array and update \(j\) downward so \(f[j-1]\) is still the old value. Extra space becomes \(O(k)\).

We notice that \(f[i][j]\) is only related to \(f[i - 1][j - 1]\) and \(f[i - 1][j]\), so we can use a one-dimensional array to optimize the space complexity.

The time complexity is \(O(nk)\), and the space complexity is \(O(k)\), where \(n\) and \(k\) are the two integers given in the problem.

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class Solution:
    def rearrangeSticks(self, n: int, k: int) -> int:
        mod = 10**9 + 7
        f = [1] + [0] * k
        for i in range(1, n + 1):
            for j in range(k, 0, -1):
                f[j] = (f[j] * (i - 1) + f[j - 1]) % mod
            f[0] = 0
        return f[k]

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class Solution {
    public int rearrangeSticks(int n, int k) {
        final int mod = (int) 1e9 + 7;
        int[] f = new int[k + 1];
        f[0] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = k; j > 0; --j) {
                f[j] = (int) ((f[j] * (i - 1L) + f[j - 1]) % mod);
            }
            f[0] = 0;
        }
        return f[k];
    }
}

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class Solution {
public:
    int rearrangeSticks(int n, int k) {
        const int mod = 1e9 + 7;
        int f[k + 1];
        memset(f, 0, sizeof(f));
        f[0] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = k; j; --j) {
                f[j] = (f[j - 1] + f[j] * (i - 1LL)) % mod;
            }
            f[0] = 0;
        }
        return f[k];
    }
};

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func rearrangeSticks(n int, k int) int {
    const mod = 1e9 + 7
    f := make([]int, k+1)
    f[0] = 1
    for i := 1; i <= n; i++ {
        for j := k; j > 0; j-- {
            f[j] = (f[j-1] + f[j]*(i-1)) % mod
        }
        f[0] = 0
    }
    return f[k]
}

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function rearrangeSticks(n: number, k: number): number {
    const mod = 10 ** 9 + 7;
    const f: number[] = Array(n + 1).fill(0);
    f[0] = 1;
    for (let i = 1; i <= n; ++i) {
        for (let j = k; j; --j) {
            f[j] = (f[j] * (i - 1) + f[j - 1]) % mod;
        }
        f[0] = 0;
    }
    return f[k];
}

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