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
const closestCost = function (baseCosts, toppingCosts, target) {
    let closestDessertCost = -Infinity;
    function dfs(dessertCost, j) {
        const tarCurrDiff = Math.abs(target - dessertCost);
        const tarCloseDiff = Math.abs(target - closestDessertCost);
        if (tarCurrDiff < tarCloseDiff) {
            closestDessertCost = dessertCost;
        } else if (tarCurrDiff === tarCloseDiff && dessertCost < closestDessertCost) {
            closestDessertCost = dessertCost;
        }
        if (dessertCost > target) return;
        if (j === toppingCosts.length) return;
        for (let count = 0; count <= 2; count++) {
            dfs(dessertCost + count * toppingCosts[j], j + 1);
        }
    }
    for (let i = 0; i < baseCosts.length; i++) {
        dfs(baseCosts[i], 0);
    }
    return closestDessertCost;
};
