Drop Eggs

There is a building ofnfloors. If an egg drops from the_k_th floor or above, it will break. If it's dropped from any floor below, it will not break.

You're given two eggs, Find_k_while minimize the number of drops for the worst case. Return the number of drops in the worst case.

Clarification

For n = 10, a naive way to find_k_is drop egg from 1st floor, 2nd floor ... kth floor. But in this worst case (k = 10), you have to drop 10 times.

Notice that you have two eggs, so you can drop at 4th, 7th & 9th floor, in the worst case (for example, k = 9) you have to drop 4 times.

Example

Given n =10, return4.
Given n =100, return14.

Solution 1: Similar to 2, only consider 1 dimension, Memory Limit Exceed

public int dropEggs(int n) {
    if(n <= 0) {
        return 0;
    }
    int[][] dp = new int[3][n + 1];
    for(int i = 0; i <= 2; i++) {
        dp[i][0] = 0;
        dp[i][1] = 1;
    }
    for(int i = 0; i <= n; i++) {
        dp[0][i] = 0;
        dp[1][i] = i;
    }
    for(int i = 2; i <= 2; i++) {
        for(int j = 2; j <= n; j++) {
            dp[i][j] = Integer.MAX_VALUE;
            for(int floor = 1; floor <= j; floor++) {
                int worst = 1 + Math.max(dp[i - 1][floor - 1], dp[i][j - floor]);
                if(dp[i][j] > worst) {
                    dp[i][j] = worst;
                }
            }
        }
    }
    return dp[2][n];
}

Solution 2: A bit tricky. http://www.jiuzhang.com/qa/2711/

public int dropEggs(int n) {
    if(n <= 0) {
        return 0;
    }
    int i = 0, res = 0;
    long sum = 0;
    while(sum < n) {
        i++;
        res++;
        sum += i;
    }
    return res;
}

Drop Egg II

There is a building ofnfloors. If an egg drops from thekth floor or above, it will break. If it's dropped from any floor below, it will not break.

You're givenmeggs, Find k while minimize the number of drops for the worst case. Return the number of drops in the worst case.

Have you met this question in a real interview?

Yes

Example

Givenm=2,n=100return14
Givenm=2,n=36return8

Sotution 1: The problem can be divide to two subproplems:

1. If egg in floor x break, then we need try from x - 1 floor with m - 1 eggs.
2. If egg in floor x not break, then we need try from x + 1 floor with m eggs.

We can implement this by recursion. This is time consuming if the floor is high. TLE.

public int dropEggs2(int m, int n) {
    if(m == 1) {
        return n;
    }
    if(n == 0 || n == 1) {
        return n;
    }
    int res = Integer_MAX_VALUE;
    for(int i = 2; i <= n; i++) {
        //floor i consume 1, remember add it.
        int worst = 1 + Math.max(dropEggs2(m - 1, i - 1), dropEggs2(m, n - i));
        if (worst < res) {
            res = worst;
        }
    }
    return res;
}

Solution 2: DP. Reduce time complexity.

public int dropEggs2(int m, int n) {
    if(m <= 0 || n <= 0) {
        return 0;
    }
    int[][] dp = new int[m + 1][n + 1];
    for(int i = 0; i <= m; i++) {
        dp[i][0] = 0;
        dp[i][1] = 1;
    }
    for(int i = 0; i <= n; i++) {
        dp[0][i] = 0;
        dp[1][i] = i;
    }
    for(int i = 2; i <= m; i++) {
        for(int j = 2; j <= n; j++) {
            dp[i][j] = Integer.MAX_VALUE;
            int worst = 0;
            for(int floor = 1; floor <= j; floor++) {
                worst = 1 + Math.max(dp[i - 1][floor - 1], dp[i][j - floor]);
                if(worst < dp[i][j]) {
                    dp[i][j] = worst;
                }
            }
        }
    }
    return dp[m][n];
}