903. Valid Permutations for DI Sequence

标签：character   code   hat   rms   ISE   problem   java   incr   output   

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 #1: DP.[C++]

class Solution {
public:
    int numPermsDISequence(string S) {
        int n = S.length(), mod = 1e9 + 7;
        vector<vector<int>> dp(n+1, vector<int>(n+1));
        for (int j = 0; j <= n; ++j) dp[0][j] = 1;
        for (int i = 0; i < n; ++i) {
            if (S[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];
    }
};

Analysis:

I feel this code is right, but I can‘t express why.

https://leetcode.com/problems/valid-permutations-for-di-sequence/discuss/168278/C%2B%2BJavaPython-DP-Solution-O(N2)

903. Valid Permutations for DI Sequence

标签：character   code   hat   rms   ISE   problem   java   incr   output   

原文地址：https://www.cnblogs.com/ruruozhenhao/p/10340249.html