Last Stone Weight II

Problem (restated)

Smash two stones a,b: if a≠b a stone of |a-b| remains. Minimize the final stone weight (0 if none).

Intuition

Equivalent to partition stones into two piles with sums as equal as possible; answer is |sum-2S| for best subset sum S ≤ sum/2.

Approaches

export function lastStoneWeightII(stones: number[]): number {
  const total = stones.reduce((a, b) => a + b, 0);
  const target = Math.floor(total / 2);
  const dp = Array(target + 1).fill(false);
  dp[0] = true;
  for (const x of stones) {
    for (let c = target; c >= x; c--) dp[c] = dp[c] || dp[c - x];
  }
  for (let s = target; s >= 0; s--) if (dp[s]) return total - 2 * s;
  return total;
}

export function lastStoneWeightII(stones: number[]): number {
  const total = stones.reduce((a, b) => a + b, 0);
  const target = Math.floor(total / 2);
  const dp = Array(target + 1).fill(false);
  dp[0] = true;
  for (const x of stones) {
    for (let c = target; c >= x; c--) dp[c] = dp[c] || dp[c - x];
  }
  for (let s = target; s >= 0; s--) if (dp[s]) return total - 2 * s;
  return total;
}

Template connection

Subset-sum / partition DP.

Reflection

• The leftover of two piles is sum - 2 * closest, where closest is the subset sum nearest to half the total.
• This is 0/1 subset sum, so the capacity loop walks backward. One stone is itself. Two equal stones answer 0.
• The answer is the weight gap between two groups. It is not the order in which stones are smashed.