FastPrepSpend it All

Spend it All

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Problem statement

There is a budget and an array of n costs. Repeat the following process until budget is less than the minimum element in cost.

Process

Start at element 0 and work to n - 1. At each element cost[i]:

  • If cost[i] <= budget, reduce budget by cost[i] and move to the next index. This is a purchase.
  • If cost[i] > budget, move to the next index.

The array is circular, so the next index after n - 1 is index 0. Continue until there is not enough budget to make a purchase.

Determine how many purchases are made.

Complete the function countPurchases in the editor below.

countPurchases has the following parameters:

  • int cost[n]: the costs of all the items
  • long budget: the starting amount of the budget to be spent
  • long: the number of purchases

Function

countPurchases(cost: int[], budget: long) → int

Examples

Example 1

cost = [5, 8, 3]budget = 12return = 3

Process:

  • Buy item 0 for 5 -> budget = 12 - 5 = 7.
  • Item 1 is too expensive, budget = 7.
  • Buy item 2 for 3, budget = 7 - 3 = 4.
  • Items 0 and 1 are too expensive, budget = 4.
  • Buy item 2 for 3, budget = 4 - 3 = 1.
  • Now budget = 1, and there are no more items that can be purchased.
  • Return 3, the number of items purchased.

Constraints

  • 1 <= n <= 2*105
  • 1 <= cost[i] <= 109
  • 1 <= budget <= 105

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public int countPurchases(int[] cost, long budget) {
    // write your code here
}
cost[5, 8, 3]
budget12
expected3
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