Egg Drop With 2 Eggs and N Floors

<p>You are given <strong>two identical</strong> eggs and you have access to a building with <code>n</code> floors labeled from <code>1</code> to <code>n</code>.</p> <p>You know that there exists a floor <code>f</code> where <code>0 &lt;= f &lt;= n</code> such that any egg dropped at a floor <strong>higher</strong> than <code>f</code> will <strong>break</strong>, and any egg dropped <strong>at or below</strong> floor <code>f</code> will <strong>not break</strong>.</p> <p>In each move, you may take an <strong>unbroken</strong> egg and drop it from any floor <code>x</code> (where <code>1 &lt;= x &lt;= n</code>). If the egg breaks, you can no longer use it. However, if the egg does not break, you may <strong>reuse</strong> it in future moves.</p> <p>Return <em>the <strong>minimum number of moves</strong> that you need to determine <strong>with certainty</strong> what the value of </em><code>f</code> is.</p> <p>&nbsp;</p> <p><strong class="example">Example 1:</strong></p> <pre> <strong>Input:</strong> n = 2 <strong>Output:</strong> 2 <strong>Explanation:</strong> 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&#39;t, we know that f = 1. Otherwise, if both eggs survive, we know that f = 2. </pre> <p><strong class="example">Example 2:</strong></p> <pre> <strong>Input:</strong> n = 100 <strong>Output:</strong> 14 <strong>Explanation:</strong> 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. </pre> <p>&nbsp;</p> <p><strong>Constraints:</strong></p> <ul> <li><code>1 &lt;= n &lt;= 1000</code></li> </ul>

class Solution:
    def twoEggDrop(self, n: int) -> int:
        # dp[i][j] represents the minimum number of moves needed to find the critical floor
        # with i eggs and j floors.
        # dp[i][j] = min(1 + max(dp[i - 1][x - 1], dp[i][j - x])) for x in range(1, j + 1)

        # Optimization: Instead of using a 2D DP array, we can use a mathematical approach.
        # Let's say we drop the first egg at floor x1. If it breaks, we have x1 - 1 floors left
        # and 1 egg. In the worst case, we need x1 - 1 more moves, so a total of x1 moves.
        # If it doesn't break, we have n - x1 floors left and 2 eggs. We can drop the egg
        # at floor x1 + x2. If it breaks, we need x2 - 1 more moves, so a total of 2 moves + x2 - 1
        # drops. If it doesn't break, we can drop at x1 + x2 + x3, and so on.

        # We need to find the minimum m such that x1 + x2 + ... + xm >= n, where xi represents
        # the number of floors we can check with one egg.

        # x1 + x2 + ... + xm = m + (m - 1) + (m - 2) + ... + 1 = m * (m + 1) / 2 >= n

        # m * (m + 1) >= 2 * n

        # We can iterate through m and find the smallest one that satisfies the condition.
        m = 0
        while m * (m + 1) // 2 < n:
            m += 1
        return m

Explanation:

We're trying to find the minimum number of moves (m) to determine the critical floor (f) with two eggs. The key insight is to understand the relationship between the number of moves and the maximum number of floors we can cover.

Mathematical Approach:

1. We can approach this problem with a mathematical formula rather than dynamic programming to optimize the solution.
2. The problem boils down to finding the smallest m such that the sum of the first m integers is greater than or equal to n.
3. The sum of the first m integers is given by the formula: m * (m + 1) / 2.
4. We need to find the minimum m such that m * (m + 1) / 2 >= n.

Algorithm:

1. Initialize m to 0.
2. Increment m until m * (m + 1) / 2 >= n.
3. Return m.

Example:

For n = 100:

• We need to find the smallest m such that m * (m + 1) / 2 >= 100.
• Trying different values of m:
  • If m = 10, 10 * 11 / 2 = 55 < 100
  • If m = 11, 11 * 12 / 2 = 66 < 100
  • If m = 12, 12 * 13 / 2 = 78 < 100
  • If m = 13, 13 * 14 / 2 = 91 < 100
  • If m = 14, 14 * 15 / 2 = 105 >= 100
• So, the minimum number of moves is 14.

Complexity Analysis:

• Time Complexity: O(sqrt(n)) - The while loop iterates until m * (m + 1) / 2 >= n. Since m is approximately the square root of n, the time complexity is O(sqrt(n)).
• Space Complexity: O(1) - We only use a constant amount of extra space.

Edge Cases:

• n = 1: The algorithm will return 1, which is correct.
• n = 2: The algorithm will return 2, which is correct.
• n = 1000: The algorithm will return 45, which is correct.

Alternative Dynamic Programming Approach (Less Efficient):

A naive approach would be to use dynamic programming. However, this leads to increased time complexity.

class Solution:
    def twoEggDrop(self, n: int) -> int:
        dp = {}  # Memoization for (eggs, floors)

        def solve(eggs, floors):
            if floors <= 1:
                return floors
            if eggs == 1:
                return floors

            if (eggs, floors) in dp:
                return dp[(eggs, floors)]

            min_moves = float('inf')
            for x in range(1, floors + 1):
                # Egg breaks at floor x: 1 move + solve(eggs - 1, x - 1) floors below
                # Egg doesn't break: 1 move + solve(eggs, floors - x) floors above
                moves = 1 + max(solve(eggs - 1, x - 1), solve(eggs, floors - x))
                min_moves = min(min_moves, moves)

            dp[(eggs, floors)] = min_moves
            return min_moves

        return solve(2, n)

Dynamic Programming Explanation:

• dp[i][j] stores the minimum number of moves needed with i eggs and j floors.
• The base cases are when there's only one egg or when there's one or zero floors. If there is only 1 floor, the answer is one. If there are zero floors, the answer is zero.
• For each floor x we try to drop the egg at, we take the worst-case scenario between the egg breaking and not breaking, and add 1 (for the current drop).
• We minimize over all possible x to get the optimal solution.

Dynamic Programming Complexity:

• Time Complexity: O(n^2) with memoization. Without memoization is exponential.
• Space Complexity: O(n) for the memoization table (dictionary).