Sponsored
Sponsored
Use these hints if you're stuck. Try solving on your own first.
Consider infected children as <code>0</code> and non-infected as <code>1</code>, then divide the array into segments with the same value.
For each segment of non-infected children whose indices are <code>[i, j]</code> and indices <code>(i - 1)</code> and <code>(j + 1)</code>, if they exist, are already infected. Then if <code>i == 0</code> or <code>j == n - 1</code>, each second there is only one kid that can be infected (which is at the other endpoint).
If <code>i > 0</code> and <code>j < n - 1</code>, we have two choices per second since the children at the two endpoints can both be the infect candidates. So there are <code>2<sup>j - i</sup></code> orders to infect all children in the segment.
Each second we can select a segment and select one endpoint from it.
The answer is: <code>S! / (len[1]! * len[2]! * ... * len[m]! * len<sub>start</sub>! * len<sub>end</sub>!) * 2<sup>k</sup></code> where <code>len[1], len[2], ..., len[m]</code> are the lengths of each segment of non-infected children that have an infected child at both endpoints, <code>len<sub>start</sub></code> and <code>len<sub>end</sub></code> denote the number of non-infected children with infected child at one endpoint, <code>S</code> is the total length of all segments of non-infected children, and <code>k = (len[1] - 1) + (len[2] - 1) + ... + (len[m] - 1)</code>.