LeetCode: Closest Dessert Cost Solution

Approach

Exhausive search through all possibilities of topping

(0, 1, 2)
for each

Implementation

1function recursion(cost, toppingCosts, toppingCostsIdx, target, ref) {
2 if (toppingCostsIdx === toppingCosts.length) {
3 const diff = Math.abs(target - cost)
4 if (diff <= ref.minDiff) {
5 ref.cost = diff === ref.minDiff ? Math.min(ref.cost, cost) : cost
6 ref.minDiff = diff
7 }
8 } else {
9 recursion(
10 cost + toppingCosts[toppingCostsIdx] * 0,
11 toppingCosts,
12 toppingCostsIdx + 1,
13 target,
14 ref
15 )
16 recursion(
17 cost + toppingCosts[toppingCostsIdx] * 1,
18 toppingCosts,
19 toppingCostsIdx + 1,
20 target,
21 ref
22 )
23 recursion(
24 cost + toppingCosts[toppingCostsIdx] * 2,
25 toppingCosts,
26 toppingCostsIdx + 1,
27 target,
28 ref
29 )
30 }
31}
32
33/**
34 * @param {number[]} baseCosts
35 * @param {number[]} toppingCosts
36 * @param {number} target
37 * @return {number}
38 */
39var closestCost = function (baseCosts, toppingCosts, target) {
40 const ref = { minDiff: Infinity, cost: Infinity }
41 for (const baseCost of baseCosts) {
42 recursion(baseCost, toppingCosts, 0, target, ref)
43 }
44 return ref.cost
45}

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Tags

leetcode

recursion

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