1774. Closest Dessert Cost
Explanation:
To solve this problem, we can use a recursive approach to try all possible combinations of ice cream base flavors and toppings. We will iterate through each ice cream base flavor and for each flavor, we will try all possible combinations of toppings (including not selecting any topping). We will keep track of the closest cost to the target and return the closest cost found.
Detailed Steps:
- Define a recursive function that takes the current cost, index of the current base flavor, and index of the current topping as parameters.
- In the recursive function, iterate through all possible combinations of toppings (including not selecting any topping) and calculate the total cost.
- Update the closest cost found so far.
- Base cases:
- If the total cost equals the target, return the target as the closest cost.
- If the total cost is greater than the target or we have considered all toppings, return the current cost as the closest cost.
- Recur by trying the next topping with the same base flavor and also by moving to the next base flavor.
- Return the closest cost found.
Time Complexity:
The time complexity of this approach is O(2^m * n) where n is the number of ice cream base flavors and m is the number of types of toppings.
Space Complexity:
The space complexity of this approach is O(m) for the recursive stack.
:
class Solution {
public int closestCost(int[] baseCosts, int[] toppingCosts, int target) {
int closestCost = Integer.MAX_VALUE;
for (int baseCost : baseCosts) {
closestCost = Math.min(closestCost, calculateCost(baseCost, 0, toppingCosts, target));
}
return closestCost;
}
private int calculateCost(int currentCost, int toppingIndex, int[] toppingCosts, int target) {
if (toppingIndex == toppingCosts.length) {
return currentCost;
}
int cost1 = calculateCost(currentCost, toppingIndex + 1, toppingCosts, target);
int cost2 = calculateCost(currentCost + toppingCosts[toppingIndex], toppingIndex + 1, toppingCosts, target);
int cost3 = calculateCost(currentCost + 2 * toppingCosts[toppingIndex], toppingIndex + 1, toppingCosts, target);
int closestCost = Math.abs(cost1 - target) < Math.abs(cost2 - target) ? cost1 : cost2;
closestCost = Math.abs(closestCost - target) < Math.abs(cost3 - target) ? closestCost : cost3;
return closestCost;
}
}Code Editor (Testing phase)
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