g0801_0900.s0887_super_egg_drop.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0801_0900.s0887_super_egg_drop;

// #Hard #Dynamic_Programming #Math #Binary_Search
// #2022_03_28_Time_0_ms_(100.00%)_Space_38.7_MB_(99.74%)

/**
 * 887 - Super Egg Drop\.
 *
 * Hard
 *
 * You are given `k` identical eggs and you have access to a building with `n` floors labeled from `1` to `n`.
 *
 * You know that there exists a 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**.
 *
 * Each move, you may take an unbroken egg and drop it from any floor `x` (where `1 <= x <= n`). If the egg breaks, you can no longer use it. However, if the egg does not break, you may **reuse** it in future moves.
 *
 * Return _the **minimum number of moves** that you need to determine **with certainty** what the value of_ `f` is.
 *
 * **Example 1:**
 *
 * **Input:** k = 1, n = 2
 *
 * **Output:** 2
 *
 * **Explanation:**
 *
 * Drop the egg from floor 1. If it breaks, we know that f = 0.
 *
 * Otherwise, drop the egg from floor 2.
 *
 * If it breaks, we know that f = 1. If it does not break, then we know f = 2.
 *
 * Hence, we need at minimum 2 moves to determine with certainty what the value of f is. 
 *
 * **Example 2:**
 *
 * **Input:** k = 2, n = 6
 *
 * **Output:** 3 
 *
 * **Example 3:**
 *
 * **Input:** k = 3, n = 14
 *
 * **Output:** 4 
 *
 * **Constraints:**
 *
 * *   `1 <= k <= 100`
 * *   1 <= n <= 10⁴
**/
public class Solution {
    public int superEggDrop(int k, int n) {
        int e = k;
        int f = n;
        int[] dp = new int[e + 1];
        int counter = 1;
        while (true) {
            int temp = dp[0];
            for (int i = 1; i < dp.length; i++) {
                int val = dp[i] + temp + 1;
                temp = dp[i];
                dp[i] = val;
                if (val >= f) {
                    return counter;
                }
            }
            counter = counter + 1;
        }
    }
}