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903. Valid Permutations for DI Sequence

Description

You are given a string s of length n where s[i] is either:

• 'D' means decreasing, or
• 'I' means increasing.

A permutation perm of n + 1 integers of all the integers in the range [0, n] is called a valid permutation if for all valid i:

• If s[i] == 'D', then perm[i] > perm[i + 1], and
• If s[i] == 'I', then perm[i] < perm[i + 1].

Return the number of valid permutations perm. Since the answer may be large, return it modulo 109 + 7.

 

Example 1:

Input: s = "DID"
Output: 5
Explanation: The 5 valid permutations of (0, 1, 2, 3) are:
(1, 0, 3, 2)
(2, 0, 3, 1)
(2, 1, 3, 0)
(3, 0, 2, 1)
(3, 1, 2, 0)

Example 2:

Input: s = "D"
Output: 1

 

Constraints:

• n == s.length
• 1 <= n <= 200
• s[i] is either 'I' or 'D'.

Solutions

Solution 1: Dynamic Programming

We define $f[i][j]$ as the number of permutations that satisfy the problem’s requirements with the first $i$ characters of the string ending with the number $j$. Initially, $f[0][0]=1$, and the rest $f[0][j]=0$. The answer is $\sum_{j=0}^n f[n][j]$.

Consider $f[i][j]$, where $j \in [0, i]$.

If the $i$th character $s[i-1]$ is 'D', then $f[i][j]$ can be transferred from $f[i-1][k]$, where $k \in [j+1, i]$. Since $k-1$ can only be up to $i-1$, we move $k$ one place to the left, so $k \in [j, i-1]$. Therefore, we have $f[i][j] = \sum_{k=j}^{i-1} f[i-1][k]$.

If the $i$th character $s[i-1]$ is 'I', then $f[i][j]$ can be transferred from $f[i-1][k]$, where $k \in [0, j-1]$. Therefore, we have $f[i][j] = \sum_{k=0}^{j-1} f[i-1][k]$.

The final answer is $\sum_{j=0}^n f[n][j]$.

The time complexity is $O(n^3)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of the string.

We can optimize the time complexity to $O(n^2)$ using prefix sums. Additionally, we can optimize the space complexity to $O(n)$ using a rolling array.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  };
  

• 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
  
  

• 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;
  }
  
  

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