887. Super Egg Drop

Approach 1: Naive DP (TLE)

• Time: $O(KN^2)$
• Space: $O(KN)$

C++Java

class Solution {
 public:
  int superEggDrop(int k, int n) {
    // dp[k][n] := min # of moves to know f with k eggs and n floors
    dp.resize(k + 1, vector<int>(n + 1, -1));
    return drop(k, n);
  }

 private:
  vector<vector<int>> dp;

  int drop(int k, int n) {
    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 (dp[k][n] != -1)
      return dp[k][n];

    dp[k][n] = INT_MAX;

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

    return dp[k][n];
  }
};

class Solution {
  public int superEggDrop(int k, int n) {
    // dp[k][n] := min # of moves to know f with k eggs and n floors
    dp = new int[k + 1][n + 1];
    Arrays.stream(dp).forEach(row -> Arrays.fill(row, -1));
    return drop(k, n);
  }

  private int[][] dp;

  private int drop(int k, int n) {
    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 (dp[k][n] != -1)
      return dp[k][n];

    dp[k][n] = Integer.MAX_VALUE;

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

    return dp[k][n];
  }
}

Approach 2: DP + Binary Search

• Time: $O(KN\log N)$
• Space: $O(KN)$

C++Java

class Solution {
 public:
  int superEggDrop(int k, int n) {
    // dp[k][n] := min # of moves to know f with k eggs and n floors
    dp.resize(k + 1, vector<int>(n + 1, -1));
    return drop(k, n);
  }

 private:
  vector<vector<int>> dp;

  int drop(int k, int n) {
    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 (dp[k][n] != -1)
      return dp[k][n];

    //   broken[i] := drop(k - 1, i - 1) is increasing w/ i
    // unbroken[i] := drop(k,     n - i) is decreasing w/ i
    // dp[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);
      const int unbroken = drop(k, n - m);
      if (broken >= unbroken)
        r = m;
      else
        l = m + 1;
    }

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

class Solution {
  public int superEggDrop(int k, int n) {
    // dp[k][n] := min # of moves to know f with k eggs and n floors
    dp = new int[k + 1][n + 1];
    Arrays.stream(dp).forEach(A -> Arrays.fill(A, -1));
    return drop(k, n);
  }

  private int[][] dp;

  private int drop(int k, int n) {
    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 (dp[k][n] != -1)
      return dp[k][n];

    //   broken[i] := drop(k - 1, i - 1) is increasing w/ i
    // unbroken[i] := drop(k,     n - i) is decreasing w/ i
    // dp[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) {
      final int m = (l + r) / 2;
      final int broken = drop(k - 1, m - 1);
      final int unbroken = drop(k, n - m);
      if (broken >= unbroken)
        r = m;
      else
        l = m + 1;
    }

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

Approach 3: Genius

• Time: $O(K\log N)$
• Space: $O(KN)$

C++JavaPython

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;
  }
};

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;
  }
}

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