887. Super Egg Drop

Description

You are given K eggs, and you have access to a building with N floors from 1 to N.

Each egg is identical in function, and if an egg breaks, you cannot drop it again.

You know that there exists a floor F with 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.

Each move, you may take an egg (if you have an unbroken one) and drop it from any floor X (with 1 <= X <= N).

Your goal is to know with certainty what the value of F is.

What is the minimum number of moves that you need to know with certainty what F is, regardless of the initial value of F?

Example 1:

Input: K = 1, N = 2
Output: 2
Explanation:
Drop the egg from floor 1. If it breaks, we know with certainty that F = 0.
Otherwise, drop the egg from floor 2. If it breaks, we know with certainty that F = 1.
If it didn’t break, then we know with certainty F = 2.
Hence, we needed 2 moves in the worst case to know what F is with certainty.
Example 2:

Input: K = 2, N = 6
Output: 3
Example 3:

Input: K = 3, N = 14
Output: 4

Note:

1 <= K <= 100
1 <= N <= 10000

Problem URL

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Solution

给K个鸡蛋和N层楼，找到第F层，使得在这一层或者一下的楼层，鸡蛋摔不碎。问最坏寻找的次数。

This is problem using dynamic programming and binary search. So we get a dp[][] array first. Then use a helper function to perform binary search. If N <= 1 or K==1, we need N steps in worst case to find the floor. Then set low and high to perform binary search. Left part using K - 1 eggs in mid -1 floow. Right part using K eggs in N - mid floor. Choose the part needs more floor to determine how many times we need next. After the binary search, set result to dp and return it.

Code

1. dp solution 1

class Solution {
    public int superEggDrop(int K, int N) {
        int[][] dp = new int[K + 1][N + 1];
        return helper(K, N, dp);
    }
    
    private int helper(int K, int N, int[][] dp){
        if (N <= 1 || K == 1){
            return N;
        }
        if (dp[K][N] > 0){
            return dp[K][N];
        }
        
        int low = 1, high = N, result = N;
        while (low < high){
            int mid = (low + high) / 2;
            int left = helper(K - 1, mid - 1, dp);
            int right = helper(K, N - mid, dp);
            result = Math.min(result, Math.max(left, right) + 1);
            if (left < right){
                low = mid + 1;
            }
            else if (left > right){
                high = mid;
            }
            else{
                break;
            }
        }
        dp[K][N] = result;
        return result;
    }
}

Time Complexity: O(KNlogN)
Space Complexity: O(KN)

2. dp solution 2
   Consider this problem in a different way:
   dp[M][K]means that, given K eggs and M moves,
   what is the maximum number of floor that we can check.

The dp equation is:
dp[m][k] = dp[m - 1][k - 1] + dp[m - 1][k] + 1,
which means we take 1 move to a floor,
if egg breaks, then we can check dp[m - 1][k - 1] floors.
if egg doesn’t breaks, then we can check dp[m - 1][k] floors.

dp[m][k] is similar to the number of combinations and it increase exponentially to N

class Solution {
    public int superEggDrop(int K, int N) {
        int[][] dp = new int[N + 1][K + 1];
        int res = 0;
        while (dp[res][K] < N){
            res++;
            for (int i = 1; i <= K; i++){
                dp[res][i] = dp[res - 1][i - 1] + dp[res - 1][i] + 1;
            }
        }
        return res;
    }
}

Time Complexity: O(KNlogN)
Space Complexity: O(KN)

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Review

A O(KN^2) TLE approach which helps understanding this problem.

dp[K][N] = 1 + max(dp[K - 1][i - 1],dp[K][N - i])

class Solution {
    public int superEggDrop(int K, int N) {
        int[][] dp = new int[K + 1][N + 1];
        return helper(K, N, dp);
    }
    
    private int helper(int K, int N, int[][] dp){
        if (N <= 1 || K == 1){
            return N;
        }
        if (dp[K][N] > 0){
            return dp[K][N];
        }
        
        int min = N;
        for (int i = 1; i <= N; i++){
            int left = helper(K - 1, i - 1, dp);
            int right = helper(K, N - i, dp);
            min = Math.min(min, Math.max(left, right) + 1);
        }
        dp[K][N] = min;
        return min;
    }
}