<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>

class Solution:

    def minSteps(self, n: int) -> int:
        adjacentes = {}
        antecedentes = {}
        visitas = {}
        fila = []
        fila.append((1,0))
        visitas[(1,0)] = True
        antecedentes[(1,0)] = (-1,-1)
        while fila!=[]:
            atual = fila.pop(0)
            if atual[0]==n:
                contador = 0
                while antecedentes.get(atual) != (-1,-1):
                    atual = antecedentes[atual]
                    contador += 1
                return contador
            adjacentes[atual]=[]
            if atual[0] <= n:
                adjacentes[atual].append((atual[0],atual[0]))
            if atual[0]+atual[1] <= n:
                adjacentes[atual].append((atual[0]+atual[1],atual[1]))
            for x in adjacentes[atual]:
                if visitas.get(x, False) == False:
                    visitas[x]=True
                    antecedentes[x]=atual
                    fila.append(x)

        return 0

19006