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.

Python3JavaC++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

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;
}

We can optimize the time complexity to \(O(n^2)\) using prefix sums.

Python3JavaC++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):
            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;
}

Additionally, we can optimize the space complexity to \(O(n)\) using a rolling array.

Python3JavaC++GoTypeScript

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;
}

Comments