Closest Dessert Cost
<p>You would like to make dessert and are preparing to buy the ingredients. You have <code>n</code> ice cream base flavors and <code>m</code> types of toppings to choose from. You must follow these rules when making your dessert:</p>

<ul>
	<li>There must be <strong>exactly one</strong> ice cream base.</li>
	<li>You can add <strong>one or more</strong> types of topping or have no toppings at all.</li>
	<li>There are <strong>at most two</strong> of <strong>each type</strong> of topping.</li>
</ul>

<p>You are given three inputs:</p>

<ul>
	<li><code>baseCosts</code>, an integer array of length <code>n</code>, where each <code>baseCosts[i]</code> represents the price of the <code>i<sup>th</sup></code> ice cream base flavor.</li>
	<li><code>toppingCosts</code>, an integer array of length <code>m</code>, where each <code>toppingCosts[i]</code> is the price of <strong>one</strong> of the <code>i<sup>th</sup></code> topping.</li>
	<li><code>target</code>, an integer representing your target price for dessert.</li>
</ul>

<p>You want to make a dessert with a total cost as close to <code>target</code> as possible.</p>

<p>Return <em>the closest possible cost of the dessert to </em><code>target</code>. If there are multiple, return <em>the <strong>lower</strong> one.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> baseCosts = [1,7], toppingCosts = [3,4], target = 10
<strong>Output:</strong> 10
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> baseCosts = [2,3], toppingCosts = [4,5,100], target = 18
<strong>Output:</strong> 17
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> baseCosts = [3,10], toppingCosts = [2,5], target = 9
<strong>Output:</strong> 8
<strong>Explanation:</strong> It is possible to make desserts with cost 8 and 10. Return 8 as it is the lower cost.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == baseCosts.length</code></li>
	<li><code>m == toppingCosts.length</code></li>
	<li><code>1 &lt;= n, m &lt;= 10</code></li>
	<li><code>1 &lt;= baseCosts[i], toppingCosts[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= target &lt;= 10<sup>4</sup></code></li>
</ul>
Array; Dynamic Programming; Backtracking
func closestCost(baseCosts []int, toppingCosts []int, target int) int {
	arr := []int{}
	var dfs func(int, int)
	dfs = func(i, t int) {
		if i >= len(toppingCosts) {
			arr = append(arr, t)
			return
		}
		dfs(i+1, t)
		dfs(i+1, t+toppingCosts[i])
	}
	dfs(0, 0)
	sort.Ints(arr)
	const inf = 1 << 30
	ans, d := inf, inf
	// 选择一种冰激淋基料
	for _, x := range baseCosts {
		// 枚举子集和
		for _, y := range arr {
			// 二分查找
			i := sort.SearchInts(arr, target-x-y)
			for j := i - 1; j < i+1; j++ {
				if j >= 0 && j < len(arr) {
					t := abs(x + y + arr[j] - target)
					if d > t || (d == t && ans > x+y+arr[j]) {
						d = t
						ans = x + y + arr[j]
					}
				}
			}
		}
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}