feat: solve No.1359,1774

Author: BaffinLee

1301-1400/1359. Count All Valid Pickup and Delivery Options.md

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+# 1359. Count All Valid Pickup and Delivery Options
+
+- Difficulty: Hard.
+- Related Topics: Math, Dynamic Programming, Combinatorics.
+- Similar Questions: .
+
+## Problem
+
+Given `n` orders, each order consist in pickup and delivery services. 
+
+Count all valid pickup/delivery possible sequences such that delivery(i) is always after of pickup(i). 
+
+Since the answer may be too large, return it modulo 10^9 + 7.
+
+ 
+Example 1:
+
+```
+Input: n = 1
+Output: 1
+Explanation: Unique order (P1, D1), Delivery 1 always is after of Pickup 1.
+```
+
+Example 2:
+
+```
+Input: n = 2
+Output: 6
+Explanation: All possible orders: 
+(P1,P2,D1,D2), (P1,P2,D2,D1), (P1,D1,P2,D2), (P2,P1,D1,D2), (P2,P1,D2,D1) and (P2,D2,P1,D1).
+This is an invalid order (P1,D2,P2,D1) because Pickup 2 is after of Delivery 2.
+```
+
+Example 3:
+
+```
+Input: n = 3
+Output: 90
+```
+
+ 
+**Constraints:**
+
+
+	
+- `1 <= n <= 500`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number} n
+ * @return {number}
+ */
+var countOrders = function(n) {
+    return helper(n, 0, 0, {});
+};
+
+var helper = function(n, x, y, dp) {
+    if (x === n && y === n) return 1;
+    var mod = Math.pow(10, 9) + 7;
+    var key = `${x}-${y}`;
+    if (dp[key] === undefined) {
+        var choosePickup = x < n ? ((n - x) * helper(n, x + 1, y, dp) % mod) : 0;
+        var chooseDelivery = y < n && x > y ? ((x - y) * helper(n, x, y + 1, dp) % mod) : 0;
+        dp[key] = (choosePickup + chooseDelivery) % mod;
+    }
+    return dp[key];
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(n).

1701-1800/1774. Closest Dessert Cost.md

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+# 1774. Closest Dessert Cost
+
+- Difficulty: Medium.
+- Related Topics: Array, Dynamic Programming, Backtracking.
+- Similar Questions: .
+
+## Problem
+
+You would like to make dessert and are preparing to buy the ingredients. You have `n` ice cream base flavors and `m` types of toppings to choose from. You must follow these rules when making your dessert:
+
+
+	
+- There must be **exactly one** ice cream base.
+	
+- You can add **one or more** types of topping or have no toppings at all.
+	
+- There are **at most two** of **each type** of topping.
+
+
+You are given three inputs:
+
+
+	
+- `baseCosts`, an integer array of length `n`, where each `baseCosts[i]` represents the price of the `ith` ice cream base flavor.
+	
+- `toppingCosts`, an integer array of length `m`, where each `toppingCosts[i]` is the price of **one** of the `ith` topping.
+	
+- `target`, an integer representing your target price for dessert.
+
+
+You want to make a dessert with a total cost as close to `target` as possible.
+
+Return **the closest possible cost of the dessert to **`target`. If there are multiple, return **the **lower** one.**
+
+ 
+Example 1:
+
+```
+Input: baseCosts = [1,7], toppingCosts = [3,4], target = 10
+Output: 10
+Explanation: Consider the following combination (all 0-indexed):
+- Choose base 1: cost 7
+- Take 1 of topping 0: cost 1 x 3 = 3
+- Take 0 of topping 1: cost 0 x 4 = 0
+Total: 7 + 3 + 0 = 10.
+```
+
+Example 2:
+
+```
+Input: baseCosts = [2,3], toppingCosts = [4,5,100], target = 18
+Output: 17
+Explanation: Consider the following combination (all 0-indexed):
+- Choose base 1: cost 3
+- Take 1 of topping 0: cost 1 x 4 = 4
+- Take 2 of topping 1: cost 2 x 5 = 10
+- Take 0 of topping 2: cost 0 x 100 = 0
+Total: 3 + 4 + 10 + 0 = 17. You cannot make a dessert with a total cost of 18.
+```
+
+Example 3:
+
+```
+Input: baseCosts = [3,10], toppingCosts = [2,5], target = 9
+Output: 8
+Explanation: It is possible to make desserts with cost 8 and 10. Return 8 as it is the lower cost.
+```
+
+ 
+**Constraints:**
+
+
+	
+- `n == baseCosts.length`
+	
+- `m == toppingCosts.length`
+	
+- `1 <= n, m <= 10`
+	
+- `1 <= baseCosts[i], toppingCosts[i] <= 104`
+	
+- `1 <= target <= 104`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} baseCosts
+ * @param {number[]} toppingCosts
+ * @param {number} target
+ * @return {number}
+ */
+var closestCost = function(baseCosts, toppingCosts, target) {
+    var res = Number.MAX_SAFE_INTEGER;
+    for (var i = 0; i < baseCosts.length; i++) {
+        res = closest(target, res, baseCosts[i] + helper(toppingCosts, target - baseCosts[i], 0));
+    }
+    return res;
+};
+
+var helper = function(toppingCosts, target, i) {
+    if (i === toppingCosts.length) return 0;
+    if (target <= 0) return 0;
+    var res = Number.MAX_SAFE_INTEGER;
+    res = closest(target, res, helper(toppingCosts, target, i + 1));
+    res = closest(target, res, toppingCosts[i] + helper(toppingCosts, target - toppingCosts[i], i + 1));
+    res = closest(target, res, toppingCosts[i] * 2 + helper(toppingCosts, target - toppingCosts[i] * 2, i + 1));
+    return res;
+};
+
+var closest = function(target, num1, num2) {
+    var diff1 = Math.abs(num1 - target);
+    var diff2 = Math.abs(num2 - target);
+    if (diff1 === diff2) return Math.min(num1, num2);
+    return diff1 < diff2 ? num1 : num2;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n * m ^ 3).
+* Space complexity : O(m).