Egg Drop With 2 Eggs and N Floors

• Create a DP array dp of size n + 1, where dp[i] will store the minimum moves for i floors.
• The base case is dp[0] = 0.
• Loop for i from 1 to n to compute dp[i] for each number of floors.
• Inside this loop, initialize dp[i] to a large value.
• Start another nested loop for x from 1 to i. x represents the floor from which we drop the first egg.
• For each x, calculate the number of moves required in the worst-case scenario:
  • If the egg breaks: x moves are needed in total (1 for this drop + x-1 for linear scan with the second egg).
  • If the egg doesn't break: 1 + dp[i-x] moves are needed (1 for this drop + dp[i-x] for the remaining problem with i-x floors).
  • The worst case for a given x is max(x, 1 + dp[i-x]).
• Update dp[i] with the minimum worst-case found so far: dp[i] = min(dp[i], max(x, 1 + dp[i-x])).
• After the loops complete, dp[n] holds the final answer.

We define dp[i] as the minimum number of moves for i floors. To find dp[i], we consider dropping the first egg from a floor x (where 1 <= x <= i).

There are two outcomes:

1. Egg breaks: We have one egg left and x-1 floors below to check. With one egg, we must check each floor one by one from the bottom. This will take x-1 more moves. The total moves in this case is 1 (for the current drop) + (x-1) = x.
2. Egg does not break: We still have two eggs, and we need to solve the problem for the i-x floors above floor x. This is a subproblem that requires dp[i-x] moves. The total moves in this case is 1 (for the current drop) + dp[i-x].

Since we need to find f with certainty, we must prepare for the worst-case scenario. For a given x, the number of moves is max(x, 1 + dp[i-x]). We want to choose the floor x that minimizes this worst-case number of moves.

The recurrence relation is: dp[i] = min_{1 <= x <= i} (max(x, 1 + dp[i-x])).

We can compute dp[i] for i = 1, 2, ..., n by iterating through all possible values of x for each i. The base case is dp[0] = 0. The final answer is dp[n].

1class Solution {2    public int twoEggDrop(int n) {3        // dp[i] stores the minimum moves for i floors.4        int[] dp = new int[n + 1];5        6        // Iterate through the number of floors from 1 to n.7        for (int i = 1; i <= n; i++) {8            dp[i] = Integer.MAX_VALUE;9            // Try dropping the first egg from every floor x from 1 to i.10            for (int x = 1; x <= i; x++) {11                // Case 1: Egg breaks at floor x.12                // We have 1 egg left and x-1 floors below to check.13                // This requires x-1 more linear checks. Total moves = 1 + (x-1) = x.14                int movesIfBreaks = x;15                16                // Case 2: Egg doesn't break at floor x.17                // We have 2 eggs left and i-x floors above to check.18                // This is the subproblem for i-x floors, which takes dp[i-x] moves.19                // Total moves = 1 + dp[i-x].20                int movesIfNotBreaks = 1 + dp[i - x];21                22                // The number of moves for a given x is the maximum of the two cases (worst-case).23                int worstCaseForX = Math.max(movesIfBreaks, movesIfNotBreaks);24                25                // We choose the floor x that minimizes this worst-case.26                dp[i] = Math.min(dp[i], worstCaseForX);27            }28        }29        return dp[n];30    }31}

• Initialize moves = 0 and floorsCovered = 0.
• Start a loop that continues as long as floorsCovered < n.
• Inside the loop, increment moves by 1.
• Add the new moves to floorsCovered. This represents the new range of floors that can be checked with the second egg if the first one breaks at the current step.
• Once floorsCovered is greater than or equal to n, the loop terminates.
• Return moves.

Let's analyze the maximum number of floors, F, we can check with k moves. For our first drop (move 1), we drop an egg from floor x.

• If it breaks, we have k-1 moves and 1 egg left. With 1 egg, we can check k-1 floors sequentially. So, x-1 must be at most k-1, meaning x <= k.
• If it doesn't break, we have k-1 moves and 2 eggs left. We can check F(k-1) more floors above x. To maximize the total floors, we should choose x=k. This way, the total floors covered is F(k) = (floors below) + (current floor) + (floors above) = (k-1) + 1 + F(k-1) = k + F(k-1).

This gives a recurrence F(k) = k + F(k-1). With the base case F(1) = 1, we can see that F(k) = k + (k-1) + ... + 1 = k * (k+1) / 2.

So, the problem is to find the smallest integer k such that k * (k+1) / 2 >= n. We can find this k by starting with k=1 and incrementing it, while keeping track of the total floors covered (k*(k+1)/2).

1class Solution {2    public int twoEggDrop(int n) {3        int moves = 0;4        int floorsCovered = 0;5        while (floorsCovered < n) {6            moves++;7            floorsCovered += moves;8        }9        return moves;10    }11}

• The problem is to find the smallest integer k such that k * (k+1) / 2 >= n.
• Rewrite this as a quadratic inequality: k^2 + k - 2n >= 0.
• Solve the corresponding equation k^2 + k - 2n = 0 for k using the quadratic formula.
• The positive root is k = (-1 + sqrt(1 + 8n)) / 2.
• Since k must be an integer, take the ceiling of this result.
• Return the final integer value.

As established in the iterative approach, the problem is equivalent to finding the smallest integer k that satisfies the inequality: k * (k+1) / 2 >= n.

This can be rewritten as a quadratic inequality: k^2 + k >= 2n, or k^2 + k - 2n >= 0.

To find the boundary for k, we can solve the corresponding quadratic equation: k^2 + k - 2n = 0. Using the quadratic formula, k = (-b ± sqrt(b^2 - 4ac)) / 2a, with a=1, b=1, c=-2n: k = (-1 ± sqrt(1^2 - 4 * 1 * (-2n))) / 2 k = (-1 ± sqrt(1 + 8n)) / 2

Since the number of moves k must be positive, we take the positive root: k = (-1 + sqrt(1 + 8n)) / 2

The solution k must be an integer. Since we are looking for the smallest integer k that satisfies the inequality, we need to take the ceiling of the result from the formula. This gives us a direct formula to calculate the result.

1class Solution {2    public int twoEggDrop(int n) {3        // We need to find the smallest k such that k * (k + 1) / 2 >= n.4        // This is equivalent to k^2 + k - 2n >= 0.5        // Solving for k in k^2 + k - 2n = 0 using the quadratic formula:6        // k = (-1 + sqrt(1 + 8n)) / 27        // Since k must be an integer, we take the ceiling of this value.8        9        double k = (Math.sqrt(1 + 8.0 * n) - 1) / 2.0;10        return (int) Math.ceil(k);11    }12}