1884. Egg Drop With 2 Eggs and N Floors

Leetcode Link

Problem Explanation:

In this problem, we are given two identical eggs and a building with n floors (numbered 1 to n). We must find the 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. The goal is to find the minimum number of moves required to determine the value of f with certainty. To solve this problem, we'll be using dynamic programming.

Let's work through an example:

Example

1
2
3Input: n = 2
4Output: 2

For this example, we have 2 floors. We can drop the first egg from floor 1 and the second egg from floor 2:

  1. If the first egg breaks, we know that f = 0.
  2. If the second egg breaks but the first egg didn’t, we know that f = 1.
  3. Otherwise, if both eggs survive, we know that f = 2.

We only needed 2 moves to determine the value of f, so the output is 2.

Solution Approach

The solution approach for this problem includes using dynamic programming and a 2D array dp with n + 1 rows and 2 columns, where dp[i][j] represents the minimum number of moves required to determine the value of f with i floors and j + 1 eggs. We will loop over all possible values of j and get the minimum possible value of dp[i][1].

Algorithm Steps

  1. Create a 2D array dp with n+1 rows and 2 columns.
  2. Set dp[i][0] = i for all 0 <= i <= n.
  3. Set dp[1][1] = 1 for a single floor.
  4. For 2 <= i <= n, if the previous egg was dropped from floor j (1 <= j < i), then dp[i][1] = max(dp[j - 1][0], dp[i - j][1]) + 1.
  5. Loop over all possible values of j and get the minimum possible value of dp[i][1].
  6. Finally, return dp[n][1].

Python Solution

1
2python
3class Solution:   
4    def twoEggDrop(self, n: int) -> int:
5        def drop(k: int, N: int) -> int:
6            if k == 0:
7                return 0
8            if k == 1:
9                return N
10            if N == 0:
11                return 0
12            if N == 1:
13                return 1
14            if dp[k][N] != -1:
15                return dp[k][N]
16            
17            l, r = 1, N + 1
18
19            while l < r:
20                m = (l + r) // 2
21                broken = drop(k - 1, m - 1)
22                unbroken = drop(k, N - m)
23                if broken >= unbroken:
24                    r = m
25                else:
26                    l = m + 1
27
28            dp[k][N] = 1 + drop(k - 1, l - 1)
29            return dp[k][N]
30
31        dp = [[-1] * (n + 1) for _ in range(3)]
32        return drop(2, n)

Java Solution

1
2java
3class Solution {
4    public int twoEggDrop(int n) {
5        int[][] dp = new int[3][n + 1];
6
7        for (int i = 0; i <= n; i++) {
8            dp[0][i] = 0;
9            dp[1][i] = i;
10            dp[2][i] = -1;
11        }
12
13        for (int i = 1; i <= n; i++) {
14            int minMoves = Integer.MAX_VALUE;
15            for (int j = 1; j < i; j++) {
16                int moves = Math.max(dp[0][j - 1], dp[1][i - j]) + 1;
17                minMoves = Math.min(minMoves, moves);
18            }
19            dp[2][i] = minMoves;
20
21            if (dp[1][i] == dp[2][i]) {
22                break;
23            }
24
25            dp[1][i] = dp[2][i];
26        }
27
28        return dp[2][n];
29    }
30}

JavaScript Solution

1
2javascript
3class Solution {
4    twoEggDrop(n) {
5        const dp = Array.from({length: 3}, () => Array(n + 1).fill(-1));
6
7        for (let i = 0; i <= n; i++) {
8            dp[0][i] = 0;
9            dp[1][i] = i;
10        }
11
12        for (let i = 1; i <= n; i++) {
13            dp[2][i] = dp[1][i];
14            for (let j = 1; j < i; j++) {
15                const moves = Math.max(dp[0][j - 1], dp[1][i - j]) + 1;
16                dp[2][i] = Math.min(dp[2][i], moves);
17            }
18
19            if (dp[1][i] === dp[2][i]) {
20                break;
21            }
22
23            dp[1][i] = dp[2][i];
24        }
25
26        return dp[2][n];
27    }
28}

C++ Solution

1
2cpp
3class Solution {
4public:
5    int twoEggDrop(int n) {
6        vector<vector<int>> dp(3, vector<int>(n + 1, -1));
7
8        for (int i = 0; i <= n; i++) {
9            dp[0][i] = 0;
10            dp[1][i] = i;
11        }
12
13        for (int i = 1; i <= n; i++) {
14            dp[2][i] = dp[1][i];
15            for (int j = 1; j < i; j++) {
16                int moves = max(dp[0][j - 1], dp[1][i - j]) + 1;
17                dp[2][i] = min(dp[2][i], moves);
18            }
19
20            if (dp[1][i] == dp[2][i]) {
21                break;
22            }
23
24            dp[1][i] = dp[2][i];
25        }
26
27        return dp[2][n];
28    }
29};

C# Solution

1
2csharp
3public class Solution {
4    public int TwoEggDrop(int n) {
5        int[][] dp = new int[3][];
6        for (int i = 0; i < 3; i++) {
7            dp[i] = new int[n + 1];
8        }
9
10        for (int i = 0; i <= n; i++) {
11            dp[0][i] = 0;
12            dp[1][i] = i;
13            dp[2][i] = -1;
14        }
15
16        for (int i = 1; i <= n; i++) {
17            int minMoves = int.MaxValue;
18            for (int j = 1; j < i; j++) {
19                int moves = Math.Max(dp[0][j - 1], dp[1][i - j]) + 1;
20                minMoves = Math.Min(minMoves, moves);
21            }
22            dp[2][i] = minMoves;
23
24            if (dp[1][i] == dp[2][i]) {
25                break;
26            }
27
28            dp[1][i] = dp[2][i];
29        }
30
31        return dp[2][n];
32    }
33}

In the solutions above, the logic behind the code in each language is explained step-by-step in the algorithm steps section, as we perform dynamic programming to build up the dp array and find the minimum number of moves required to determine the value of f.The solutions above for Python, Java, JavaScript, C++, and C# define the standard dynamic programming method for solving the egg drop problem with two eggs and n floors. These solutions have a time complexity of O(n^2) and O(n) space complexity. You can try implementing the given solutions in your preferred programming language to understand how to solve this problem effectively.

By following the algorithm steps mentioned above and analyzing the code provided for each language, we can better understand the process of problem-solving using dynamic programming. This helps us break down complex problems into simpler sub-problems and solve them in a systematic manner for efficient solutions. Besides, practicing with multiple languages will help you become a better programmer in general, as it strengthens your ability to adapt and learn different programming paradigms.

In conclusion, the egg drop problem is a classic question in dynamic programming that can be solved using various techniques. The methods provided above are well-explained and optimized to solve this problem for given constraints. Implementing the solutions in different programming languages will help you enhance your programming skills and strengthen your understanding of dynamic programming concepts.

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Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

Which one best describes the time complexity of the following code?

1int factorial(int n) {
2  if (n < 0) {
3    return -1;
4  } else if (n == 0) {
5    return 1;
6  } else {
7    return n * factorial(n - 1);
8  }
9}

Solution Implementation

Not Sure What to Study? Take the 2-min Quiz：

What is the best way of checking if an element exists in an unsorted array once in terms of time complexity? Select the best that applies.

Fast Track Your Learning with Our Quick Skills Quiz:

Given a sorted array of integers and an integer called target, find the element that equals to the target and return its index. Select the correct code that fills the ___ in the given code snippet.

1def binary_search(arr, target):
2    left, right = 0, len(arr) - 1
3    while left ___ right:
4        mid = (left + right) // 2
5        if arr[mid] == target:
6            return mid
7        if arr[mid] < target:
8            ___ = mid + 1
9        else:
10            ___ = mid - 1
11    return -1
12
1public static int binarySearch(int[] arr, int target) {
2    int left = 0;
3    int right = arr.length - 1;
4
5    while (left ___ right) {
6        int mid = left + (right - left) / 2;
7        if (arr[mid] == target) return mid;
8        if (arr[mid] < target) {
9            ___ = mid + 1;
10        } else {
11            ___ = mid - 1;
12        }
13    }
14    return -1;
15}
16
1function binarySearch(arr, target) {
2    let left = 0;
3    let right = arr.length - 1;
4
5    while (left ___ right) {
6        let mid = left + Math.trunc((right - left) / 2);
7        if (arr[mid] == target) return mid;
8        if (arr[mid] < target) {
9            ___ = mid + 1;
10        } else {
11            ___ = mid - 1;
12        }
13    }
14    return -1;
15}
16

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