1884. Egg Drop With 2 Eggs and N Floors LeetCode Solution

Last updated: February 25, 2025

In this guide, you will get 1884. Egg Drop With 2 Eggs and N Floors LeetCode Solution with the best time and space complexity. The solution to Egg Drop With Eggs and N Floors problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Egg Drop With Eggs and N Floors solution in C++
Egg Drop With Eggs and N Floors solution in Java
Egg Drop With Eggs and N Floors solution in Python
Additional Resources

1884. Egg Drop With 2 Eggs and N Floors LeetCode Solution image

Problem Statement of Egg Drop With Eggs and N Floors

You are given two identical eggs and you have access to a building with n floors labeled from 1 to n.
You know that there exists a floor f where 0 <= f <= n such that any egg dropped at a floor higher than f will break, and any egg dropped at or below floor f will not break.
In each move, you may take an unbroken egg and drop it from any floor x (where 1 <= x <= n). If the egg breaks, you can no longer use it. However, if the egg does not break, you may reuse it in future moves.
Return the minimum number of moves that you need to determine with certainty what the value of f is.

Example 1:

Input: n = 2
Output: 2
Explanation: We can drop the first egg from floor 1 and the second egg from floor 2.
If the first egg breaks, we know that f = 0.
If the second egg breaks but the first egg didn’t, we know that f = 1.
Otherwise, if both eggs survive, we know that f = 2.

Example 2:

Input: n = 100
Output: 14
Explanation: One optimal strategy is:
– Drop the 1st egg at floor 9. If it breaks, we know f is between 0 and 8. Drop the 2nd egg starting from floor 1 and going up one at a time to find f within 8 more drops. Total drops is 1 + 8 = 9.
– If the 1st egg does not break, drop the 1st egg again at floor 22. If it breaks, we know f is between 9 and 21. Drop the 2nd egg starting from floor 10 and going up one at a time to find f within 12 more drops. Total drops is 2 + 12 = 14.
– If the 1st egg does not break again, follow a similar process dropping the 1st egg from floors 34, 45, 55, 64, 72, 79, 85, 90, 94, 97, 99, and 100.
Regardless of the outcome, it takes at most 14 drops to determine f.

Constraints:

1 <= n <= 1000

Complexity Analysis

Time Complexity: O(n\log n)
Space Complexity: O(n)

1884. Egg Drop With 2 Eggs and N Floors LeetCode Solution in C++

class Solution {
 public:
  int twoEggDrop(int n) {
    return superEggDrop(2, n);
  }

 private:
  // Same as 887. Super Egg Drop
  int superEggDrop(int k, int n) {
    vector<vector<int>> mem(k + 1, vector<int>(n + 1, -1));
    return drop(k, n, mem);
  }

  // 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);
  }
};
/* code provided by PROGIEZ */

1884. Egg Drop With 2 Eggs and N Floors LeetCode Solution in Java

class Solution {
  public int twoEggDrop(int n) {
    return superEggDrop(2, n);
  }

  // Same as 887. Super Egg Drop
  private 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);
  }
}
// code provided by PROGIEZ

1884. Egg Drop With 2 Eggs and N Floors LeetCode Solution in Python

N/A
 # code by PROGIEZ

Additional Resources

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