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| 1 | +/* |
| 2 | + * @lc app=leetcode id=1359 lang=cpp |
| 3 | + * |
| 4 | + * [1359] Count All Valid Pickup and Delivery Options |
| 5 | + * |
| 6 | + * https://leetcode.com/problems/count-all-valid-pickup-and-delivery-options/description/ |
| 7 | + * |
| 8 | + * algorithms |
| 9 | + * Hard (58.25%) |
| 10 | + * Likes: 81 |
| 11 | + * Dislikes: 11 |
| 12 | + * Total Accepted: 3.5K |
| 13 | + * Total Submissions: 6K |
| 14 | + * Testcase Example: '1' |
| 15 | + * |
| 16 | + * Given n orders, each order consist in pickup and delivery services. |
| 17 | + * |
| 18 | + * Count all valid pickup/delivery possible sequences such that delivery(i) is |
| 19 | + * always after of pickup(i). |
| 20 | + * |
| 21 | + * Since the answer may be too large, return it modulo 10^9 + 7. |
| 22 | + * |
| 23 | + * |
| 24 | + * Example 1: |
| 25 | + * |
| 26 | + * |
| 27 | + * Input: n = 1 |
| 28 | + * Output: 1 |
| 29 | + * Explanation: Unique order (P1, D1), Delivery 1 always is after of Pickup |
| 30 | + * 1. |
| 31 | + * |
| 32 | + * |
| 33 | + * Example 2: |
| 34 | + * |
| 35 | + * |
| 36 | + * Input: n = 2 |
| 37 | + * Output: 6 |
| 38 | + * Explanation: All possible orders: |
| 39 | + * (P1,P2,D1,D2), (P1,P2,D2,D1), (P1,D1,P2,D2), (P2,P1,D1,D2), (P2,P1,D2,D1) |
| 40 | + * and (P2,D2,P1,D1). |
| 41 | + * This is an invalid order (P1,D2,P2,D1) because Pickup 2 is after of Delivery |
| 42 | + * 2. |
| 43 | + * |
| 44 | + * |
| 45 | + * Example 3: |
| 46 | + * |
| 47 | + * |
| 48 | + * Input: n = 3 |
| 49 | + * Output: 90 |
| 50 | + * |
| 51 | + * |
| 52 | + * |
| 53 | + * Constraints: |
| 54 | + * |
| 55 | + * |
| 56 | + * 1 <= n <= 500 |
| 57 | + * |
| 58 | + */ |
| 59 | + |
| 60 | +// @lc code=start |
| 61 | +class Solution { |
| 62 | +public: |
| 63 | + long long M; |
| 64 | + vector<int> dp; |
| 65 | + int f(long long n){ |
| 66 | + if(n <= 1) return 1; |
| 67 | + |
| 68 | + if(dp[n] != -1) |
| 69 | + return dp[n]; |
| 70 | + |
| 71 | + return dp[n] = n*(f(n-1) + (n-1)*(n*2-2)*(n*2-3)*f(n-2)%M)%M; |
| 72 | + } |
| 73 | + int countOrders(int n) { |
| 74 | + M = 1000000000+7; |
| 75 | + dp = vector<int>(n+1, -1); |
| 76 | + return f(n); |
| 77 | + } |
| 78 | +}; |
| 79 | +// @lc code=end |
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