LeetCode 903 - Valid Permutations for DI Sequence

https://leetcode.com/problems/valid-permutations-for-di-sequence/

We are given S, a length n string of characters from the set {'D', 'I'}. (These letters stand for "decreasing" and "increasing".)

A valid permutation is a permutation P[0], P[1], ..., P[n] of integers {0, 1, ..., n}, such that for all i:

• If S[i] == 'D', then P[i] > P[i+1], and;
• If S[i] == 'I', then P[i] < P[i+1].

How many valid permutations are there?  Since the answer may be large, return your answer modulo 10^9 + 7.

Example 1:

Input: "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)

Note:

1. 1 <= S.length <= 200
2. S consists only of characters from the set {'D', 'I'}.

Approach 2: Divide and Conquer

Let's place the zero of the permutation first. It either goes between a 'DI' part of the sequence, or it could go on the ends (the left end if it starts with 'I', and the right end if it ends in 'D'.) Afterwards, this splits the problem into two disjoint subproblems that we can solve with similar logic.

Algorithm

Let dp(i, j) be the number of valid permutations (of n = j-i+2 total integers from 0 to n-1) corresponding to the DI sequence S[i], S[i+1], ..., S[j]. If we can successfully place a zero between S[k-1] and S[k], then there are two disjoint problems S[i], ..., S[k-2] and S[k+1], ..., S[j].

To count the number of valid permutations in this case, we should choose k-i elements from n-1 (n total integers, minus the zero) to put in the left group; then the answer is this, times the number of ways to arrange the left group [dp(i, k-2)], times the number of ways to arrange the right group [dp(k+1, j)].

• Time Complexity: $O(N^2)$, where $N$ is the length of S.
• Space Complexity: $O(N^2)$. 

from functools import lru_cache
class Solution:
    def numPermsDISequence(self, S):
        MOD = 10**9 + 7

        fac = [1, 1]
        for x in range(2, 201):
            fac.append(fac[-1] * x % MOD)
        facinv = [pow(f, MOD-2, MOD) for f in fac]

        def binom(n, k):
            return fac[n] * facinv[n-k] % MOD * facinv[k] % MOD

        @lru_cache(None)
        def dp(i, j):
            if i >= j: return 1
            ans = 0
            n = j - i + 2
            if S[i] == 'I': ans += dp(i+1, j)
            if S[j] == 'D': ans += dp(i, j-1)

            for k in range(i+1, j+1):
                if S[k-1:k+1] == 'DI':
                    ans += binom(n-1, k-i) * dp(i, k-2) % MOD * dp(k+1, j) % MOD
                    ans %= MOD
            return ans

        return dp(0, len(S) - 1)
https://leetcode.com/problems/valid-permutations-for-di-sequence/discuss/168278/C%2B%2BJavaPython-DP-Solution-O(N2)

dp[i][j] means the number of possible permutations of first i + 1 digits,
where the i + 1th digit is j + 1th smallest in the rest of digits.

Ok, may not make sense ... Let's see the following diagram.

I take the example of S = "DID".
The permutation can start from 1, 2, 3, 4.
So dp[0][0] = dp[0][1] = dp[0][2] = dp[0][3] = 1.
In the parenthesis, I list all possible permutations.

We decrese from the first digit to the second,
the down arrow show the all possibile decresing pathes.

The same, cause we increase from the second digit to the third,
the up arrow show the all possibile increasing pathes.

dp[2][1] = 5, mean the number of permutations
where the third digitis the second smallest of the rest.
We have 413,314,214,423,324.
Fow example 413, where 2,3 are left and 3 the second smallest of them.

Explanation:
As shown in the diagram,
for "I", we calculate prefix sum of the array,
for "D", we calculate sufixsum of the array.

Time Complexity:
O(N^2)

    public int numPermsDISequence(String S) {
        int n = S.length(), mod = (int)1e9 + 7;
        int[][] dp = new int[n + 1][n + 1];
        for (int j = 0; j <= n; j++) dp[0][j] = 1;
        for (int i = 0; i < n; i++)
            if (S.charAt(i) == 'I')
                for (int j = 0, cur = 0; j < n - i; j++)
                    dp[i + 1][j] = cur = (cur + dp[i][j]) % mod;
            else
                for (int j = n - i - 1, cur = 0; j >= 0; j--)
                    dp[i + 1][j] = cur = (cur + dp[i][j + 1]) % mod;
        return dp[n][0];
    }

Now as we did for every DP, make it 1D dp.
Reminded by @apple702, in the Java solution, it should be dp=Arrays.copyOf(dp2, n);
Otherwise it passes an address.

    public int numPermsDISequence(String S) {
        int n = S.length(), mod = (int)1e9 + 7;
        int[] dp = new int[n + 1], dp2 = new int[n];;
        for (int j = 0; j <= n; j++) dp[j] = 1;
        for (int i = 0; i < n; i++) {
            if (S.charAt(i) == 'I')
                for (int j = 0, cur = 0; j < n - i; j++)
                    dp2[j] = cur = (cur + dp[j]) % mod;
            else
                for (int j = n - i - 1, cur = 0; j >= 0; j--)
                    dp2[j] = cur = (cur + dp[j + 1]) % mod;
            dp = Arrays.copyOf(dp2, n);
        }
        return dp[0];
    }

Approach 1: Dynamic Programming

When writing the permutation P = P_0, P_1, ..., P_N from left to right, we only care about the relative rank of the last element placed. For example, if N = 5 (so that we have elements {0, 1, 2, 3, 4, 5}), and our permutation starts 2, 3, 4, then it is similar to a situation where we have placed ?, ?, 2 and the remaining elements are {0, 1, 3}, in terms of how many possibilities there are to place the remaining elements in a valid way.

To this end, let dp(i, j) be the number of ways to place every number up to and inlcuding P_i, such that P_i when placed had relative rank j. (Namely, there are j remaining numbers less than P_i.)

Algorithm

When placing P_i following a decreasing instruction S[i-1] == 'D', we want P_{i-1} to have a higher value. When placing P_i following an increasing instruction, we want P_{i-1} to have a lower value. It is relatively easy to deduce the recursion from this fact.

  public int numPermsDISequence(String S) {

    int MOD = 1_000_000_007;

    int N = S.length();

    // dp[i][j] : Num ways to place P_i with relative rank j

    int[][] dp = new int[N + 1][N + 1];

    Arrays.fill(dp[0], 1);

    for (int i = 1; i <= N; ++i) {

      for (int j = 0; j <= i; ++j) {

        if (S.charAt(i - 1) == 'D') {

          for (int k = j; k < i; ++k) {

            dp[i][j] += dp[i - 1][k];

            dp[i][j] %= MOD;

          }

        } else {

          for (int k = 0; k < j; ++k) {

            dp[i][j] += dp[i - 1][k];

            dp[i][j] %= MOD;

          }

        }

      }

    }

    int ans = 0;

    for (int x : dp[N]) {

      ans += x;

      ans %= MOD;

    }

    return ans;

  }

Actually, we can do better than this. For any given i, let's look at how the sum of D_k = dp(i-1, k) is queried. Assuming S[i-1] == 'I', we query D_0, D_0 + D_1, D_0 + D_1 + D_2, ... etc. The case for S[i-1] == 'D' is similar.

Thus, we don't need to query the sum every time. Instead, we could use (for S[i-1] == 'I') the fact that dp(i, j) = dp(i, j-1) + dp(i-1, j-1). For S[i-1] == 'D', we have the similar fact that dp(i, j) = dp(i, j+1) + dp(i-1, j).

These two facts make the work done for each state of dp have $O(1)$ (amortized) complexity, leading to a total time complexity of $O(N^2)$ for this solution.

• Time Complexity: $O(N^3)$, where $N$ is the length of S, or $O(N^2)$ with the optimized version.
• Space Complexity: $O(N^2)$. 

    def numPermsDISequence(self, S):

        MOD = 10**9 + 7

        N = len(S)

        @lru_cache(None)

        def dp(i, j):

            # How many ways to place P_i with relative rank j?

            if not(0 <= j <= i):

                return 0

            if i == 0:

                return 1

            elif S[i-1] == 'D':

                return (dp(i, j+1) + dp(i-1, j)) % MOD

            else:

                return (dp(i, j-1) + dp(i-1, j-1)) % MOD

        return sum(dp(N, j) for j in range(N+1)) % MOD