# 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.

• 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;
}