lastStoneWeight.js

/**
 * @param {number[]} stones
 * @return {number}
 */
var lastStoneWeight = function (stones) {
  while (stones.length >= 2) {
    stones.sort((a, b) => b - a);
    if (stones[0] == stones[1]) {
      stones.shift();
      stones.shift();
    } else {
      stones[1] = stones[0] - stones[1];
      stones.shift();
    }
  }

  return stones.length ? stones[0] : 0;
};

console.log(lastStoneWeight([2, 7, 4, 1, 8, 1])); // 1

// https://leetcode.com/explore/featured/card/30-day-leetcoding-challenge/529/week-2/3297/

// We have a collection of stones, each stone has a positive integer weight.

// Each turn, we choose the two heaviest stones and smash them together.Suppose the stones have weights x and y with x <= y.The result of this smash is:

// If x == y, both stones are totally destroyed;
// If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y - x.

// At the end, there is at most 1 stone left.Return the weight of this stone(or 0 if there are no stones left.)

// Example 1:

// Input: [2, 7, 4, 1, 8, 1]
// Output: 1
// Explanation:
// We combine 7 and 8 to get 1 so the array converts to[2, 4, 1, 1, 1] then,
//     we combine 2 and 4 to get 2 so the array converts to[2, 1, 1, 1] then,
//         we combine 2 and 1 to get 1 so the array converts to[1, 1, 1] then,
//             we combine 1 and 1 to get 0 so the array converts to[1] then that's the value of last stone.

// Note:

// 1 <= stones.length <= 30
// 1 <= stones[i] <= 1000