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.length1 <= n <= 200s[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 ∑_{j=0}n f[n][j].
Consider f[i][j], where j ∈ [0, i].
If the ith character s[i-1] is 'D', then f[i][j] can be transferred from f[i-1][k], where k ∈ [j+1, i]. Since k-1 can only be up to i-1, we move k one place to the left, so k ∈ [j, i-1]. Therefore, we have f[i][j] = ∑_{k=j}i-1 f[i-1][k].
If the ith character s[i-1] is 'I', then f[i][j] can be transferred from f[i-1][k], where k ∈ [0, j-1]. Therefore, we have f[i][j] = ∑_{k=0}j-1 f[i-1][k].
The final answer is ∑_{j=0}n f[n][j].
The time complexity is O(n3), and the space complexity is O(n2). Here, n is the length of the string.
We can optimize the time complexity to O(n2) using prefix sums.
Additionally, we can optimize the space complexity to O(n) using a rolling array.
PythonJavaC++GoTypeScriptPythonJavaC++GoTypeScriptPythonJavaC++GoTypeScript
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
