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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.

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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

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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;
    }
}

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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