<p>There is only one character <code>&#39;A&#39;</code> on the screen of a notepad. You can perform one of two operations on this notepad for each step:</p>

<ul>
	<li>Copy All: You can copy all the characters present on the screen (a partial copy is not allowed).</li>
	<li>Paste: You can paste the characters which are copied last time.</li>
</ul>

<p>Given an integer <code>n</code>, return <em>the minimum number of operations to get the character</em> <code>&#39;A&#39;</code> <em>exactly</em> <code>n</code> <em>times on the screen</em>.</p>

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

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> Initially, we have one character &#39;A&#39;.
In step 1, we use Copy All operation.
In step 2, we use Paste operation to get &#39;AA&#39;.
In step 3, we use Paste operation to get &#39;AAA&#39;.
</pre>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
</ul>

#include <queue>
#include <unordered_set>

class Solution {
public:
    int minSteps(int n) {
        if (n == 1) {
            return 0;
        }

        // Queue to store the state: (count, clipboard, steps)
        std::queue<std::tuple<int, int, int>> q;
        q.push({1, 1, 1});  // Start with one 'A' on screen, one 'A' in clipboard, and 1 step

        // Set to track visited states: (count, clipboard)
        std::unordered_set<int> visited;
        visited.insert(1 * 1000 + 1);  // Unique state identifier (count * 1000 + clipboard)

        while (!q.empty()) {
            auto [count, clipboard, steps] = q.front();
            q.pop();

            // Option 1: Paste operation
            if (count + clipboard == n) {
                return steps + 1;
            }
            if (count + clipboard < n && visited.find((count + clipboard) * 1000 + clipboard) == visited.end()) {
                q.push({count + clipboard, clipboard, steps + 1});
                visited.insert((count + clipboard) * 1000 + clipboard);
            }

            // Option 2: Copy operation
            if (visited.find(count * 1000 + count) == visited.end()) {
                q.push({count, count, steps + 1});
                visited.insert(count * 1000 + count);
            }
        }

        return -1;  // Should never reach here
    }
};


// class Solution {
// public:
//     long fun(int n, int count, int last){
//         if(count == n){
//             return 0;
//         }
//         if(count > n){
//             return INT_MAX;
//         }
//         long sum1 = 1;
//         long sum2 = 1;
//         sum1 += fun(n, count + last, last);
//         sum2 += 1 + fun(n, count + count , count);
//         return min(sum1, sum2);
//     }

//     int minSteps(int n) {
//         if(n == 1){
//             return 0;
//         }
//         int ans = 0;
//         int last = 1;
//         int count = 1;

//         ans += fun(n, count, last) + 1;

//         return ans;
//     }
// };
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