1962. Remove Stones to Minimize the Total

Description

You are given a 0-indexed integer array piles, where piles[i] represents the number of stones in the ith pile, and an integer k. You should apply the following operation exactly k times:

• Choose any piles[i] and remove floor(piles[i] / 2) stones from it.

Notice that you can apply the operation on the same pile more than once.

Return the minimum possible total number of stones remaining after applying the k operations.

floor(x) is the greatest integer that is smaller than or equal to x (i.e., rounds x down).

 

Example 1:

Input: piles = [5,4,9], k = 2
Output: 12
Explanation: Steps of a possible scenario are:
- Apply the operation on pile 2. The resulting piles are [5,4,5].
- Apply the operation on pile 0. The resulting piles are [3,4,5].
The total number of stones in [3,4,5] is 12.

Example 2:

Input: piles = [4,3,6,7], k = 3
Output: 12
Explanation: Steps of a possible scenario are:
- Apply the operation on pile 2. The resulting piles are [4,3,3,7].
- Apply the operation on pile 3. The resulting piles are [4,3,3,4].
- Apply the operation on pile 0. The resulting piles are [2,3,3,4].
The total number of stones in [2,3,3,4] is 12.

 

Constraints:

• 1 <= piles.length <= 105
• 1 <= piles[i] <= 104
• 1 <= k <= 105

 

Solutions

Solution: Greedy

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} piles
 * @param {number} k
 * @return {number}
 */
const minStoneSum = function (piles, k) {
  const CORNER_PILE = 10 ** 4;
  const pileCounts = Array.from({length: CORNER_PILE + 1}).fill(0);
  let result = 0;

  for (const pile of piles) pileCounts[pile] += 1;
  for (let pile = CORNER_PILE; pile > 0; pile--) {
    while (k && pileCounts[pile]) {
      const stones = Math.floor(pile / 2);

      pileCounts[pile - stones] += 1;
      pileCounts[pile] -= 1;
      k -= 1;
    }
    if (!pileCounts[pile]) continue;
    result += pileCounts[pile] * pile;
  }
  return result;
};