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887. Super Egg Drop 👍

Approach 1: Naive DP (TLE)

  • Time: $O(kn^2)$
  • Space: $O(kn)$
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class Solution {
 public:
  int superEggDrop(int k, int n) {
    vector<vector<int>> mem(k + 1, vector<int>(n + 1, INT_MAX));
    return drop(k, n, mem);
  }

 private:
  // Returns the minimum number of moves to know f with k eggs and n floors.
  int drop(int k, int n, vector<vector<int>>& mem) {
    if (k == 0)  // no eggs -> done
      return 0;
    if (k == 1)  // one egg -> drop from 1-th floor to n-th floor
      return n;
    if (n == 0)  // no floor -> done
      return 0;
    if (n == 1)  // one floor -> drop from that floor
      return 1;
    if (mem[k][n] != INT_MAX)
      return mem[k][n];

    for (int i = 1; i <= n; ++i) {
      const int broken = drop(k - 1, i - 1, mem);
      const int unbroken = drop(k, n - i, mem);
      mem[k][n] = min(mem[k][n], 1 + max(broken, unbroken));
    }

    return mem[k][n];
  }
};

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class Solution {
  public int superEggDrop(int k, int n) {
    int[][] mem = new int[k + 1][n + 1];
    Arrays.stream(mem).forEach(A -> Arrays.fill(A, Integer.MAX_VALUE));
    return drop(k, n, mem);
  }

  // Returns the minimum number of moves to know f with k eggs and n floors.
  private int drop(int k, int n, int[][] mem) {
    if (k == 0) // no eggs -> done
      return 0;
    if (k == 1) // one egg -> drop from 1st floor to nth floor
      return n;
    if (n == 0) // no floor -> done
      return 0;
    if (n == 1) // one floor -> drop from that floor
      return 1;
    if (mem[k][n] != Integer.MAX_VALUE)
      return mem[k][n];

    for (int i = 1; i <= n; ++i) {
      final int broken = drop(k - 1, i - 1, mem);
      final int unbroken = drop(k, n - i, mem);
      mem[k][n] = Math.min(mem[k][n], 1 + Math.max(broken, unbroken));
    }

    return mem[k][n];
  }
}
  • Time: $O(kn\log n)$
  • Space: $O(kn)$
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class Solution {
 public:
  int superEggDrop(int k, int n) {
    vector<vector<int>> mem(k + 1, vector<int>(n + 1, -1));
    return drop(k, n, mem);
  }

 private:
  // Returns the minimum number of moves to know f with k eggs and n floors.
  int drop(int k, int n, vector<vector<int>>& mem) {
    if (k == 0)  // no eggs -> done
      return 0;
    if (k == 1)  // one egg -> drop from 1-th floor to n-th floor
      return n;
    if (n == 0)  // no floor -> done
      return 0;
    if (n == 1)  // one floor -> drop from that floor
      return 1;
    if (mem[k][n] != -1)
      return mem[k][n];

    //   broken[i] := drop(k - 1, i - 1) is increasing with i
    // unbroken[i] := drop(k,     n - i) is decreasing with i
    // mem[k][n] := 1 + min(max(broken[i], unbroken[i])), 1 <= i <= n
    // Find the first index i s.t broken[i] >= unbroken[i], which minimizes
    // max(broken[i], unbroken[i]).

    int l = 1;
    int r = n + 1;

    while (l < r) {
      const int m = (l + r) / 2;
      const int broken = drop(k - 1, m - 1, mem);
      const int unbroken = drop(k, n - m, mem);
      if (broken >= unbroken)
        r = m;
      else
        l = m + 1;
    }

    return mem[k][n] = 1 + drop(k - 1, l - 1, mem);
  }
};

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class Solution {
  public int superEggDrop(int k, int n) {
    int[][] mem = new int[k + 1][n + 1];
    Arrays.stream(mem).forEach(A -> Arrays.fill(A, -1));
    return drop(k, n, mem);
  }

  // Returns the minimum number of moves to know f with k eggs and n floors.
  private int drop(int k, int n, int[][] mem) {
    if (k == 0) // no eggs -> done
      return 0;
    if (k == 1) // one egg -> drop from 1-th floor to n-th floor
      return n;
    if (n == 0) // no floor -> done
      return 0;
    if (n == 1) // one floor -> drop from that floor
      return 1;
    if (mem[k][n] != -1)
      return mem[k][n];

    int l = 1;
    int r = n + 1;

    while (l < r) {
      final int m = (l + r) / 2;
      final int broken = drop(k - 1, m - 1, mem);
      final int unbroken = drop(k, n - m, mem);
      if (broken >= unbroken)
        r = m;
      else
        l = m + 1;
    }

    return mem[k][n] = 1 + drop(k - 1, l - 1, mem);
  }
}

Approach 3: Genius

  • Time: $O(k\log n)$
  • Space: $O(kn)$
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class Solution {
 public:
  int superEggDrop(int k, int n) {
    int moves = 0;
    vector<vector<int>> dp(n + 1, vector<int>(k + 1));

    while (dp[moves][k] < n) {
      ++moves;
      for (int eggs = 1; eggs <= k; ++eggs)
        dp[moves][eggs] = dp[moves - 1][eggs - 1] + dp[moves - 1][eggs] + 1;
    }

    return moves;
  }
};

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class Solution {
  public int superEggDrop(int k, int n) {
    int moves = 0;
    int[][] dp = new int[n + 1][k + 1];

    while (dp[moves][k] < n) {
      ++moves;
      for (int eggs = 1; eggs <= k; ++eggs)
        dp[moves][eggs] = dp[moves - 1][eggs - 1] + dp[moves - 1][eggs] + 1;
    }

    return moves;
  }
}

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class Solution:
  def superEggDrop(self, k: int, n: int) -> int:
    moves = 0
    dp = [[0] * (k + 1) for _ in range(n + 1)]

    while dp[moves][k] < n:
      moves += 1
      for eggs in range(1, k + 1):
        dp[moves][eggs] = (dp[moves - 1][eggs - 1] +
                           dp[moves - 1][eggs] + 1)

    return moves