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1049. Last Stone Weight II

medium

Smash stones into two groups; minimize the final residue. Reduces to: pick a subset closest to half the total — knapsack feasibility maximized.

By Sam K., Founder, InterviewChamp.AI · Last verified

Problem

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 any two 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 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 smallest possible weight of the left stone. If there are no stones left, return 0.

Constraints

  • 1 <= stones.length <= 30
  • 1 <= stones[i] <= 100

Examples

Example 1

Input
stones = [2,7,4,1,8,1]
Output
1

Explanation: We can combine 2 and 4 to get 2, so the array converts to [2,7,1,8,1] then, combine 7 and 8 to get 1, so the array converts to [2,1,1,1] then, combine 2 and 1 to get 1, so the array converts to [1,1,1] then, combine 1 and 1 to get 0, so the array converts to [1], then that's the optimal value.

Example 2

Input
stones = [31,26,33,21,40]
Output
5

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Output

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Hints

Progressive — try the first before opening the next.

Hint 1

Equivalent: assign each stone a + or - so the resulting absolute sum is minimal.

Hint 2

If P is the sum of the + group and N of the - group, total = P + N and answer = |P - N| = total - 2P.

Hint 3

Minimize that by maximizing P with P <= total/2. Subset-sum knapsack on the array.

Solution approach

Reveal approach

Any sequence of smashes corresponds to assigning each stone a + or - sign and the final residue equals |sum(+) - sum(-)|. To minimize the residue, minimize total - 2*P where P is the subset sum of the + group, constrained to P <= total/2. Solve via boolean knapsack: dp[s] is whether some subset sums to s, for s up to total/2. Initialize dp[0] = true. For each stone w, sweep s from total/2 down to w and set dp[s] |= dp[s - w]. Scan dp from total/2 downward to find the largest reachable P. Return total - 2*P. O(n * total) time, O(total) space.

Complexity

Time
O(n * sum)
Space
O(sum)

Related patterns

  • dynamic-programming
  • knapsack-0-1

Related problems

Asked at

Companies reported asking this problem (sourced from public Glassdoor, Blind, and Levels.fyi interview posts).

  • Amazon

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