You are given an array of integers stones where stones[i] is the weight of the ith stone.

We are playing a game with the stones. On each turn, we choose the heaviest two stones and smash them together. Suppose the heaviest two stones have weights x and y with x <= y. The result of this smash is:

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

At the end of the game, there is at most one stone left.

Return the weight of the last remaining stone. If there are no stones left, return 0.

Example

Input: stones = [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 the last stone.

Solution

Great recursive solution found here.

/**
 * @param {number[]} stones
 * @return {number}
 */
var lastStoneWeight = function (stones) {
  //If there is only one stone left, return it
  if (stones.length < 2) return stones;
  //Sort the stones in ascending order
  stones.sort((a, b) => a - b);
  //Get the difference between the two heaviest stones
  let a = stones.pop();
  let b = stones.pop();
  //Push the difference back into the array
  stones.push(Math.abs(a - b));
  //Recursively call the function until there is only one stone left
  return lastStoneWeight(stones);
};