Question

1049. Last Stone Weight II

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Description

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.

 

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,
we can combine 7 and 8 to get 1, so the array converts to [2,1,1,1] then,
we can combine 2 and 1 to get 1, so the array converts to [1,1,1] then,
we can 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

 

Constraints:

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

Solutions

Solution 1

class Solution:
  def lastStoneWeightII(self, stones: List[int]) -> int:
    s = sum(stones)
    m, n = len(stones), s >> 1
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    for i in range(1, m + 1):
      for j in range(n + 1):
        dp[i][j] = dp[i - 1][j]
        if stones[i - 1] <= j:
          dp[i][j] = max(
            dp[i][j], dp[i - 1][j - stones[i - 1]] + stones[i - 1]
          )
    return s - 2 * dp[-1][-1]

class Solution {
  public int lastStoneWeightII(int[] stones) {
    int s = 0;
    for (int v : stones) {
      s += v;
    }
    int m = stones.length;
    int n = s >> 1;
    int[][] dp = new int[m + 1][n + 1];
    for (int i = 1; i <= m; ++i) {
      for (int j = 0; j <= n; ++j) {
        dp[i][j] = dp[i - 1][j];
        if (stones[i - 1] <= j) {
          dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - stones[i - 1]] + stones[i - 1]);
        }
      }
    }
    return s - dp[m][n] * 2;
  }
}

class Solution {
public:
  int lastStoneWeightII(vector<int>& stones) {
    int s = accumulate(stones.begin(), stones.end(), 0);
    int m = stones.size(), n = s >> 1;
    vector<vector<int>> dp(m + 1, vector<int>(n + 1));
    for (int i = 1; i <= m; ++i) {
      for (int j = 0; j <= n; ++j) {
        dp[i][j] = dp[i - 1][j];
        if (stones[i - 1] <= j) dp[i][j] = max(dp[i][j], dp[i - 1][j - stones[i - 1]] + stones[i - 1]);
      }
    }
    return s - dp[m][n] * 2;
  }
};

func lastStoneWeightII(stones []int) int {
  s := 0
  for _, v := range stones {
    s += v
  }
  m, n := len(stones), s>>1
  dp := make([][]int, m+1)
  for i := range dp {
    dp[i] = make([]int, n+1)
  }
  for i := 1; i <= m; i++ {
    for j := 0; j <= n; j++ {
      dp[i][j] = dp[i-1][j]
      if stones[i-1] <= j {
        dp[i][j] = max(dp[i][j], dp[i-1][j-stones[i-1]]+stones[i-1])
      }
    }
  }
  return s - dp[m][n]*2
}

impl Solution {
  #[allow(dead_code)]
  pub fn last_stone_weight_ii(stones: Vec<i32>) -> i32 {
    let n = stones.len();
    let mut sum = 0;

    for e in &stones {
      sum += *e;
    }

    let m = (sum / 2) as usize;
    let mut dp: Vec<Vec<i32>> = vec![vec![0; m + 1]; n + 1];

    // Begin the actual dp process
    for i in 1..=n {
      for j in 1..=m {
        dp[i][j] = if stones[i - 1] > (j as i32) {
          dp[i - 1][j]
        } else {
          std::cmp::max(
            dp[i - 1][j],
            dp[i - 1][j - (stones[i - 1] as usize)] + stones[i - 1]
          )
        };
      }
    }

    sum - 2 * dp[n][m]
  }
}

/**
 * @param {number[]} stones
 * @return {number}
 */
var lastStoneWeightII = function (stones) {
  let s = 0;
  for (let v of stones) {
    s += v;
  }
  const n = s >> 1;
  let dp = new Array(n + 1).fill(0);
  for (let v of stones) {
    for (let j = n; j >= v; --j) {
      dp[j] = Math.max(dp[j], dp[j - v] + v);
    }
  }
  return s - dp[n] * 2;
};

Solution 2

class Solution:
  def lastStoneWeightII(self, stones: List[int]) -> int:
    s = sum(stones)
    m, n = len(stones), s >> 1
    dp = [0] * (n + 1)
    for v in stones:
      for j in range(n, v - 1, -1):
        dp[j] = max(dp[j], dp[j - v] + v)
    return s - dp[-1] * 2

class Solution {
  public int lastStoneWeightII(int[] stones) {
    int s = 0;
    for (int v : stones) {
      s += v;
    }
    int m = stones.length;
    int n = s >> 1;
    int[] dp = new int[n + 1];
    for (int v : stones) {
      for (int j = n; j >= v; --j) {
        dp[j] = Math.max(dp[j], dp[j - v] + v);
      }
    }
    return s - dp[n] * 2;
  }
}

class Solution {
public:
  int lastStoneWeightII(vector<int>& stones) {
    int s = accumulate(stones.begin(), stones.end(), 0);
    int n = s >> 1;
    vector<int> dp(n + 1);
    for (int& v : stones)
      for (int j = n; j >= v; --j)
        dp[j] = max(dp[j], dp[j - v] + v);
    return s - dp[n] * 2;
  }
};

func lastStoneWeightII(stones []int) int {
  s := 0
  for _, v := range stones {
    s += v
  }
  n := s >> 1
  dp := make([]int, n+1)
  for _, v := range stones {
    for j := n; j >= v; j-- {
      dp[j] = max(dp[j], dp[j-v]+v)
    }
  }
  return s - dp[n]*2
}