Question

903. DI 序列的有效排列

English Version

题目描述

给定一个长度为 n 的字符串 s ，其中 s[i] 是:

• “D” 意味着减少，或者
• “I” 意味着增加

有效排列 是对有 n + 1 个在 [0, n]  范围内的整数的一个排列 perm ，使得对所有的 i：

• 如果 s[i] == 'D'，那么 perm[i] > perm[i+1]，以及；
• 如果 s[i] == 'I'，那么 perm[i] < perm[i+1]。

返回 _有效排列  _perm_的数量 _。因为答案可能很大，所以请返回你的答案对 109 + 7 取余。

 

示例 1：

输入：s = "DID"
输出：5
解释：
(0, 1, 2, 3) 的五个有效排列是：
(1, 0, 3, 2)
(2, 0, 3, 1)
(2, 1, 3, 0)
(3, 0, 2, 1)
(3, 1, 2, 0)

示例 2:

输入: s = "D"
输出: 1

 

提示:

• n == s.length
• 1 <= n <= 200
• s[i] 不是 'I' 就是 'D'

解法

方法一：动态规划

我们定义 $f[i][j]$ 表示字符串的前 $i$ 个字符中，以数字 $j$ 结尾的满足题目要求的排列的数量。初始时 $f[0][0]=1$，其余 $f[0][j]=0$。答案为 $\sum_{j=0}^n f[n][j]$。

考虑 $f[i][j]$，其中 $j \in [0, i]$。

如果第 $i$ 个字符 $s[i-1]$ 是 'D'，那么 $f[i][j]$ 可以从 $f[i-1][k]$ 转移而来，其中 $k \in [j+1, i]$，而由于 $k-1$ 最大只能为 $i-1$，我们将 $k$ 向左移动一位，那么 $k \in [j, i-1]$，因此有 $f[i][j] = \sum_{k=j}^{i-1} f[i-1][k]$。

如果第 $i$ 个字符 $s[i-1]$ 是 'I'，那么 $f[i][j]$ 可以从 $f[i-1][k]$ 转移而来，其中 $k \in [0, j-1]$，因此有 $f[i][j] = \sum_{k=0}^{j-1} f[i-1][k]$。

最终的答案即为 $\sum_{j=0}^n f[n][j]$。

时间复杂度 $O(n^3)$，空间复杂度 $O(n^2)$。其中 $n$ 是字符串的长度。

class Solution:
  def numPermsDISequence(self, s: str) -> int:
    mod = 10**9 + 7
    n = len(s)
    f = [[0] * (n + 1) for _ in range(n + 1)]
    f[0][0] = 1
    for i, c in enumerate(s, 1):
      if c == "D":
        for j in range(i + 1):
          for k in range(j, i):
            f[i][j] = (f[i][j] + f[i - 1][k]) % mod
      else:
        for j in range(i + 1):
          for k in range(j):
            f[i][j] = (f[i][j] + f[i - 1][k]) % mod
    return sum(f[n][j] for j in range(n + 1)) % mod

class Solution {
  public int numPermsDISequence(String s) {
    final int mod = (int) 1e9 + 7;
    int n = s.length();
    int[][] f = new int[n + 1][n + 1];
    f[0][0] = 1;
    for (int i = 1; i <= n; ++i) {
      if (s.charAt(i - 1) == 'D') {
        for (int j = 0; j <= i; ++j) {
          for (int k = j; k < i; ++k) {
            f[i][j] = (f[i][j] + f[i - 1][k]) % mod;
          }
        }
      } else {
        for (int j = 0; j <= i; ++j) {
          for (int k = 0; k < j; ++k) {
            f[i][j] = (f[i][j] + f[i - 1][k]) % mod;
          }
        }
      }
    }
    int ans = 0;
    for (int j = 0; j <= n; ++j) {
      ans = (ans + f[n][j]) % mod;
    }
    return ans;
  }
}

class Solution {
public:
  int numPermsDISequence(string s) {
    const int mod = 1e9 + 7;
    int n = s.size();
    int f[n + 1][n + 1];
    memset(f, 0, sizeof(f));
    f[0][0] = 1;
    for (int i = 1; i <= n; ++i) {
      if (s[i - 1] == 'D') {
        for (int j = 0; j <= i; ++j) {
          for (int k = j; k < i; ++k) {
            f[i][j] = (f[i][j] + f[i - 1][k]) % mod;
          }
        }
      } else {
        for (int j = 0; j <= i; ++j) {
          for (int k = 0; k < j; ++k) {
            f[i][j] = (f[i][j] + f[i - 1][k]) % mod;
          }
        }
      }
    }
    int ans = 0;
    for (int j = 0; j <= n; ++j) {
      ans = (ans + f[n][j]) % mod;
    }
    return ans;
  }
};

func numPermsDISequence(s string) (ans int) {
  const mod = 1e9 + 7
  n := len(s)
  f := make([][]int, n+1)
  for i := range f {
    f[i] = make([]int, n+1)
  }
  f[0][0] = 1
  for i := 1; i <= n; i++ {
    if s[i-1] == 'D' {
      for j := 0; j <= i; j++ {
        for k := j; k < i; k++ {
          f[i][j] = (f[i][j] + f[i-1][k]) % mod
        }
      }
    } else {
      for j := 0; j <= i; j++ {
        for k := 0; k < j; k++ {
          f[i][j] = (f[i][j] + f[i-1][k]) % mod
        }
      }
    }
  }
  for j := 0; j <= n; j++ {
    ans = (ans + f[n][j]) % mod
  }
  return
}

function numPermsDISequence(s: string): number {
  const n = s.length;
  const f: number[][] = Array(n + 1)
    .fill(0)
    .map(() => Array(n + 1).fill(0));
  f[0][0] = 1;
  const mod = 10 ** 9 + 7;
  for (let i = 1; i <= n; ++i) {
    if (s[i - 1] === 'D') {
      for (let j = 0; j <= i; ++j) {
        for (let k = j; k < i; ++k) {
          f[i][j] = (f[i][j] + f[i - 1][k]) % mod;
        }
      }
    } else {
      for (let j = 0; j <= i; ++j) {
        for (let k = 0; k < j; ++k) {
          f[i][j] = (f[i][j] + f[i - 1][k]) % mod;
        }
      }
    }
  }
  let ans = 0;
  for (let j = 0; j <= n; ++j) {
    ans = (ans + f[n][j]) % mod;
  }
  return ans;
}

我们可以用前缀和优化时间复杂度，使得时间复杂度降低到 $O(n^2)$。

class Solution:
  def numPermsDISequence(self, s: str) -> int:
    mod = 10**9 + 7
    n = len(s)
    f = [[0] * (n + 1) for _ in range(n + 1)]
    f[0][0] = 1
    for i, c in enumerate(s, 1):
      pre = 0
      if c == "D":
        for j in range(i, -1, -1):
          pre = (pre + f[i - 1][j]) % mod
          f[i][j] = pre
      else:
        for j in range(i + 1):
          f[i][j] = pre
          pre = (pre + f[i - 1][j]) % mod
    return sum(f[n][j] for j in range(n + 1)) % mod

class Solution {
  public int numPermsDISequence(String s) {
    final int mod = (int) 1e9 + 7;
    int n = s.length();
    int[][] f = new int[n + 1][n + 1];
    f[0][0] = 1;
    for (int i = 1; i <= n; ++i) {
      int pre = 0;
      if (s.charAt(i - 1) == 'D') {
        for (int j = i; j >= 0; --j) {
          pre = (pre + f[i - 1][j]) % mod;
          f[i][j] = pre;
        }
      } else {
        for (int j = 0; j <= i; ++j) {
          f[i][j] = pre;
          pre = (pre + f[i - 1][j]) % mod;
        }
      }
    }
    int ans = 0;
    for (int j = 0; j <= n; ++j) {
      ans = (ans + f[n][j]) % mod;
    }
    return ans;
  }
}

class Solution {
public:
  int numPermsDISequence(string s) {
    const int mod = 1e9 + 7;
    int n = s.size();
    int f[n + 1][n + 1];
    memset(f, 0, sizeof(f));
    f[0][0] = 1;
    for (int i = 1; i <= n; ++i) {
      int pre = 0;
      if (s[i - 1] == 'D') {
        for (int j = i; j >= 0; --j) {
          pre = (pre + f[i - 1][j]) % mod;
          f[i][j] = pre;
        }
      } else {
        for (int j = 0; j <= i; ++j) {
          f[i][j] = pre;
          pre = (pre + f[i - 1][j]) % mod;
        }
      }
    }
    int ans = 0;
    for (int j = 0; j <= n; ++j) {
      ans = (ans + f[n][j]) % mod;
    }
    return ans;
  }
};

func numPermsDISequence(s string) (ans int) {
  const mod = 1e9 + 7
  n := len(s)
  f := make([][]int, n+1)
  for i := range f {
    f[i] = make([]int, n+1)
  }
  f[0][0] = 1
  for i := 1; i <= n; i++ {
    pre := 0
    if s[i-1] == 'D' {
      for j := i; j >= 0; j-- {
        pre = (pre + f[i-1][j]) % mod
        f[i][j] = pre
      }
    } else {
      for j := 0; j <= i; j++ {
        f[i][j] = pre
        pre = (pre + f[i-1][j]) % mod
      }
    }
  }
  for j := 0; j <= n; j++ {
    ans = (ans + f[n][j]) % mod
  }
  return
}

function numPermsDISequence(s: string): number {
  const n = s.length;
  const f: number[][] = Array(n + 1)
    .fill(0)
    .map(() => Array(n + 1).fill(0));
  f[0][0] = 1;
  const mod = 10 ** 9 + 7;
  for (let i = 1; i <= n; ++i) {
    let pre = 0;
    if (s[i - 1] === 'D') {
      for (let j = i; j >= 0; --j) {
        pre = (pre + f[i - 1][j]) % mod;
        f[i][j] = pre;
      }
    } else {
      for (let j = 0; j <= i; ++j) {
        f[i][j] = pre;
        pre = (pre + f[i - 1][j]) % mod;
      }
    }
  }
  let ans = 0;
  for (let j = 0; j <= n; ++j) {
    ans = (ans + f[n][j]) % mod;
  }
  return ans;
}

另外，我们也可以用滚动数组优化空间复杂度，使得空间复杂度降低到 $O(n)$。

class Solution:
  def numPermsDISequence(self, s: str) -> int:
    mod = 10**9 + 7
    n = len(s)
    f = [1] + [0] * n
    for i, c in enumerate(s, 1):
      pre = 0
      g = [0] * (n + 1)
      if c == "D":
        for j in range(i, -1, -1):
          pre = (pre + f[j]) % mod
          g[j] = pre
      else:
        for j in range(i + 1):
          g[j] = pre
          pre = (pre + f[j]) % mod
      f = g
    return sum(f) % mod

class Solution {
  public int numPermsDISequence(String s) {
    final int mod = (int) 1e9 + 7;
    int n = s.length();
    int[] f = new int[n + 1];
    f[0] = 1;
    for (int i = 1; i <= n; ++i) {
      int pre = 0;
      int[] g = new int[n + 1];
      if (s.charAt(i - 1) == 'D') {
        for (int j = i; j >= 0; --j) {
          pre = (pre + f[j]) % mod;
          g[j] = pre;
        }
      } else {
        for (int j = 0; j <= i; ++j) {
          g[j] = pre;
          pre = (pre + f[j]) % mod;
        }
      }
      f = g;
    }
    int ans = 0;
    for (int j = 0; j <= n; ++j) {
      ans = (ans + f[j]) % mod;
    }
    return ans;
  }
}

class Solution {
public:
  int numPermsDISequence(string s) {
    const int mod = 1e9 + 7;
    int n = s.size();
    vector<int> f(n + 1);
    f[0] = 1;
    for (int i = 1; i <= n; ++i) {
      int pre = 0;
      vector<int> g(n + 1);
      if (s[i - 1] == 'D') {
        for (int j = i; j >= 0; --j) {
          pre = (pre + f[j]) % mod;
          g[j] = pre;
        }
      } else {
        for (int j = 0; j <= i; ++j) {
          g[j] = pre;
          pre = (pre + f[j]) % mod;
        }
      }
      f = move(g);
    }
    int ans = 0;
    for (int j = 0; j <= n; ++j) {
      ans = (ans + f[j]) % mod;
    }
    return ans;
  }
};

func numPermsDISequence(s string) (ans int) {
  const mod = 1e9 + 7
  n := len(s)
  f := make([]int, n+1)
  f[0] = 1
  for i := 1; i <= n; i++ {
    pre := 0
    g := make([]int, n+1)
    if s[i-1] == 'D' {
      for j := i; j >= 0; j-- {
        pre = (pre + f[j]) % mod
        g[j] = pre
      }
    } else {
      for j := 0; j <= i; j++ {
        g[j] = pre
        pre = (pre + f[j]) % mod
      }
    }
    f = g
  }
  for j := 0; j <= n; j++ {
    ans = (ans + f[j]) % mod
  }
  return
}

function numPermsDISequence(s: string): number {
  const n = s.length;
  let f: number[] = Array(n + 1).fill(0);
  f[0] = 1;
  const mod = 10 ** 9 + 7;
  for (let i = 1; i <= n; ++i) {
    let pre = 0;
    const g: number[] = Array(n + 1).fill(0);
    if (s[i - 1] === 'D') {
      for (let j = i; j >= 0; --j) {
        pre = (pre + f[j]) % mod;
        g[j] = pre;
      }
    } else {
      for (let j = 0; j <= i; ++j) {
        g[j] = pre;
        pre = (pre + f[j]) % mod;
      }
    }
    f = g;
  }
  let ans = 0;
  for (let j = 0; j <= n; ++j) {
    ans = (ans + f[j]) % mod;
  }
  return ans;
}