<p>You are given <code>k</code> identical eggs and you have access to a building with <code>n</code> floors labeled from <code>1</code> to <code>n</code>.</p>

<p>You know that there exists a floor <code>f</code> where <code>0 &lt;= f &lt;= n</code> such that any egg dropped at a floor <strong>higher</strong> than <code>f</code> will <strong>break</strong>, and any egg dropped <strong>at or below</strong> floor <code>f</code> will <strong>not break</strong>.</p>

<p>Each move, you may take an unbroken egg and drop it from any floor <code>x</code> (where <code>1 &lt;= x &lt;= n</code>). If the egg breaks, you can no longer use it. However, if the egg does not break, you may <strong>reuse</strong> it in future moves.</p>

<p>Return <em>the <strong>minimum number of moves</strong> that you need to determine <strong>with certainty</strong> what the value of </em><code>f</code> is.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> k = 1, n = 2
<strong>Output:</strong> 2
<strong>Explanation: </strong>
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> k = 2, n = 6
<strong>Output:</strong> 3
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> k = 3, n = 14
<strong>Output:</strong> 4
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= 100</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
</ul>


class Solution {
public:
    unordered_map<long long, int> memo;

    int dp(int K, int N) {
        if (K == 1) return N;
        if (N == 0) return 0;
        
        long long key = (long long)K * 100000 + N; // K와 N 조합을 키로 사용
        if (memo.find(key) != memo.end()) return memo[key];

        int low = 1, high = N, res = INT_MAX;
        while (low <= high) {
            int mid = (low + high) / 2;
            int break_count = dp(K - 1, mid - 1);  // 계란이 깨지는 경우
            int no_break_count = dp(K, N - mid);   // 계란이 깨지지 않는 경우

            if (break_count > no_break_count) {
                high = mid - 1;
                res = min(res, break_count + 1);
            } else {
                low = mid + 1;
                res = min(res, no_break_count + 1);
            }
        }

        memo[key] = res;
        return res;
    }

    int superEggDrop(int k, int n) {
        return dp(k, n);
    }
};
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