Drop Eggs

There is a building of n floors. 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, return 4. Given n = 100, return 14.

Solution:

The floor n could be represented as follows:

$$ n = x + (x - 1) + (x - 2) + ... + 1

$$ Therefore, we could sum from 1 to x, if this sum is greater than or equal to n, then x will be the minimum drops.

Code:

public int dropEggs(int n) {
    long sum = 0;
    for (long i = 1; ; i++) {
        sum += i;
        if (sum >= n) {
            return (int)i;;
        }
    }
}