103503A — Make Sum Great Again
Author: Gheal
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103503B — Binary Search Search
Author: Gheal
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Solution> Solution:</p><p>To better understand the solution, let's consider $$$n=12$$$. The final array is the BFS traversal starting from the root of the following binary tree:</p><p>\begin{center}\includegraphics[scale=0.7]{arbore_binar_fml_2.png}\end{center}</p><p>This binary tree was generated by starting with the root $$$m=\lfloor \frac{n+1}{2} \rfloor=6$$$. Each node $$$m=\lfloor \frac{l+r}{2}\rfloor$$$ has up to two children: $$$m_1=\lfloor \frac{l+(m-1)}{2} \rfloor$$$ and $$$m_2=\lfloor \frac{r+(m+1)}{2} \rfloor$$$. </p><p>If our given integer $$$k$$$ is in a complete layer, then we can employ a binary search-type algorithm to find the position of $$$k$$$:</p><p>\begin{lstlisting} l = 1, r = n, position = 1, visited_nodes = 1 while(visited_nodes<=n){ m = (l + r) / 2</p> <pre>if(k==m){ print position return } if(k<m) position = position * 2 else position = position * 2 + 1 visited_nodes = visited_nodes * 2 + 1</pre><p>} \end{lstlisting}</p><p>This algorithm has the added benefit of not printing anything if $$$k$$$ is not in a complete layer. In this case, the position of $$$k$$$ in the last layer can be determined from $$$position$$$, although this isn't as straightforward. If the binary tree were complete, then the position of $$$k$$$ in the last layer would be $$$position-\frac{\text{visited_nodes}-1}{2}$$$, since there are $$$\frac{\text{visited_nodes}-1}{2}$$$ nodes on the other layers, all of which will be printed before $$$k$$$.</p><p>By looking at the missing nodes from the binary tree for $$$n=12$$$, one may make the crucial observation that the general case position for $$$k$$$ in the last layer is equal to $$$k-(position-\frac{\text{visited_nodes}-1}{2})$$$. </p><p>In conclusion, the final position for $$$k$$$ will be $$$k-(position-\frac{\text{visited_nodes}-1}{2})+\frac{\text{visited_nodes}-1}{2}=k+\text{visited_nodes}-position-1$$$.</p><p>Time complexity per testcase: $$$O(log n)$$$ </spoiler></p>
