Google Code Jam Archive — Farewell Round C problems

Test Set 1

In Test Set 1, the initial idea is to try all possible permutations of each part, but the limits are not small enough to try everything. We observe that it is always better to sort the first part to be the (lexicographically) smallest possible permutation, and to sort the last part to be the (lexicographically) largest possible permutation. Then we can try all permutations of the middle part (there are at most $$$8!$$$) and check if any of them makes all parts sorted in non-decreasing lexicographical order.

Test Set 2

For Test Set 2 we need a different approach, as there are too many permutations per part to try. We observe that, for each part, it is best to make it the smallest possible permutation that is not smaller than the previous part (or the smallest permutation in case of the first part). By doing so, we give the next part more opportunities (if any) to be a permutation that is not smaller than the current part.

We see that the best permutation of the current part is the one that shares the longest common prefix with the previous part, such that one of the following conditions holds:

• There is a character we can append to that prefix in the current part that is larger than the next character in the previous part; or
• The whole previous part is a prefix of the current part.

One way to do this is to maintain a count of each character per part. When we are at a part, we need to find the longest common prefix with the previous part, subject to the condition above.

The solution above only requires a constant number of scans for each part. This makes the algorithm $$$O(\sum_{i} |\mathbf{S_i}|)$$$, which works for the problem constraints. However, less efficient implementations can also pass.

Test Set 1

We can simulate each query. First, we can store all the vaccines and the patients in an ordered data structure. Then, for each query, we can go through the range from start to end positions, maintain the list of picked up vaccines, and each time we encounter a patient whose vaccine is picked up and has not been administered yet, we mark them as vaccinated. This can give us the count of vaccinated patients for every move. For each query, there will be at most $$$O(\mathbf{V})$$$ patients and vaccines we will encounter. Saving the picked up vaccine and looking up if we can vaccinate someone with the available vaccines can be done in $$$O(\log\mathbf{V})$$$ using a set-like data structure. So the total time complexity of this approach is $$$O(\mathbf{M} \times \mathbf{V}\log\mathbf{V})$$$, which is sufficient for Test Set 1.

Test Set 2

For Test Set 2, we need to optimize the simulation. One important observation here is that for each patient who is located to the east side of their vaccine, they would be vaccinated when the robot is moving in the east direction. Let us consider two cases where that can happen. The vaccine is at $$$x$$$, the patient at $$$y$$$, and $$$y$$$ > $$$x$$$ (the patient is on the east of the vaccine). Let us assume the move range from $$$r_1$$$ to $$$r_2$$$.

1. The move range covers both the vaccine and the patient. The vaccine will be picked up, and then the patient will be vaccinated in the same move.
   
2. The move covers only the vaccine, not the patient. The vaccine will be picked in this move and the robot will finish at position $$$r_2$$$. To complete the vaccination, the robot needs to go to $$$y$$$, so it must move in the east direction to come to position $$$y$$$.
   

The same can be said about the patients who are on the west side of their vaccines, that they will be vaccinated when the robot is moving in the west direction. With this insight, we can process these two types of cases separately. Let us call them east case (where patients will be vaccinated in an east move) and west case (where patients will be vaccinated in a west move).

For the east cases, we can keep track of the farthest right point $$$Max_R$$$ that the robot visited so far. When the robot is making an east move with the range ($$$r_1$$$,$$$r_2$$$] and $$$r_2$$$ is on the west of the $$$Max_R$$$, then this range has been already visited by the robot, and nothing new happens. Otherwise, we can count the east case patients in that range(max($$$r_1$$$, $$$Max_R$$$), $$$r_2$$$], and that count will be the answer of the query. Since each patient can be vaccinated only once, this does not take longer than $$$O(\mathbf{V})$$$ time in total.

For the west cases, this simple strategy does not work, because we need to know not just the farthest right point, but also all the west case vaccines that have been picked in previous moves, and not used yet. Since the robot starts at position 0, and the vaccines and patients are always on the positive coordinates, for any west move that visits a patient, their vaccine is already picked up in one of the past east moves. So we can maintain an ordered list of west case patients whose vaccines have been picked during the previous east moves, and in each west move, find the number of patients in that move range from the list. We should later delete those patients. Since each patient will be counted and deleted only once, and we can complete each find, insert, delete operations in $$$O(\log\mathbf{V})$$$ time, the total time complexity here is $$$O(\mathbf{V} \log\mathbf{V})$$$. The overall time complexity of the whole solution is $$$O(\mathbf{M} + \mathbf{V} + \mathbf{V} \log\mathbf{V})$$$, which is sufficient for Test Set 2. $$$O(\mathbf{M})$$$ is included since we need to go through each query once at least.

From the definition of an interesting triplet $$$(a,b,c)$$$, it is clear that the node $$$b$$$ has a central role, so let us find the number of interesting triplets for a fixed node $$$b$$$. If $$$A_b$$$ is the number of nodes $$$a$$$ in the subtree of $$$b$$$ such that $$$\mathbf{S_b} \gt \mathbf{K} \times \mathbf{S_a}$$$ and $$$C_b$$$ is the number of nodes $$$c$$$ outside the subtree of $$$b$$$ such that $$$\mathbf{S_b} \gt \mathbf{K} \times \mathbf{S_c}$$$, then the number of interesting triplets with the given middle node $$$b$$$ is $$$A_b \times C_b$$$. The answer is the sum $$$\sum_{b=1}^\mathbf{N} A_b \times C_b$$$ of contributions over all nodes $$$b$$$.

Test Set 1

The constraint $$$\mathbf{N} \le 1000$$$ suggests that we are a looking for a quadratic algorithm. If we manage to compute $$$A_b$$$ and $$$C_b$$$ in linear time for a fixed node $$$b$$$, we would obtain a solution with the overall time complexity $$$O(\mathbf{N}^2)$$$.

For a fixed node $$$b$$$, $$$A_b$$$ can be computed by performing a Depth-first search (DFS) rooted at the node $$$b$$$ and verifying the inequality for each and every node. As for $$$C_b$$$, we can do a linear search and count the total number $$$X$$$ of nodes $$$x$$$ such that $$$\mathbf{S_b} \gt \mathbf{K} \times \mathbf{S_x}$$$ holds. Then $$$C_b=X-A_b$$$.

Test Set 2

For the large test set, the general idea remains the same. However, we cannot afford to spend linear time to compute $$$A_b$$$ and $$$C_b$$$. It seems like we need a data structure $$$D$$$ that supports the following queries efficiently:

• What is the number of nodes $$$a$$$ in the subtree rooted at $$$b$$$ such that $$$\mathbf{S_b} \gt \mathbf{K} \times \mathbf{S_a}$$$?
• What is the total number of nodes $$$x$$$ such that $$$\mathbf{S_b} \gt \mathbf{K} \times \mathbf{S_x}$$$?

That sounds complicated though. Fortunately, we can simplify these questions a great deal if we process the nodes $$$b$$$ in a non-decreasing order by $$$\mathbf{S_b}$$$. Then, by using a two pointer technique, we can make sure that the data structure $$$D$$$ contains precisely the nodes $$$x$$$ such that $$$\mathbf{S_b} \gt \mathbf{K} \times \mathbf{S_x}$$$ and nothing else. Now the above questions can be translated as follows:

• How many nodes of $$$D$$$ are in the subtree rooted at $$$b$$$?
• What is the size of $$$D$$$?

A very efficient technique for testing if a node $$$a$$$ is in the subtree of the node $$$b$$$ is called tree flattening. Essentially, we perform a post-order tree traversal to compute the labels $$$end(v)$$$ for each node $$$v$$$, which are shown at the right side of each node in the following illustration. Moreover, using the same DFS traversal, we compute the label $$$start(v)$$$ for each node $$$v$$$ (shown at the left side of nodes in the example), which is the minimum $$$end(u)$$$ over all nodes $$$u$$$ in the subtree of $$$v$$$. Then a node $$$a$$$ is in the subtree rooted at $$$b$$$ if and only if $$$start(b) \le end(a) \le end(b)$$$.

For example, the set of labels $$$end(v)$$$ of all nodes in the subtree of node $$$3$$$ in the above drawing form a consecutive interval $$$[2,6]$$$, which is conveniently stored as $$$[start(3),end(3)]$$$, hence, a node $$$v$$$ belongs to the subtree of node $$$3$$$ if and only if $$$2 \le end(v) \le 6$$$.

Answering the first question now seems a lot like a range query that a segment tree or a Fenwick tree is a perfect fit for. For example, we can use a segment tree $$$D$$$ on the range $$$[1,\mathbf{N}]$$$, which is initially empty, namely, $$$D[i]=0$$$ for all $$$i$$$. Whenever we add a node $$$v$$$ to $$$D$$$, we modify the segment tree to set $$$D[end(v)]=1$$$. It is important to note that we are marking the presence of a node at the position $$$end(v)$$$ of the segment tree rather than $$$v$$$ itself. In this way, a range query $$$D[start(b),end(b)]$$$ returns the number of nodes $$$a$$$ that are in the subtree of $$$b$$$ and present in $$$D$$$.

The time complexity of this solution is $$$O(\mathbf{N} \log \mathbf{N})$$$ because of the sorting and $$$2\mathbf{N}$$$ segment tree operations.

Let's start by treating the city as a graph, where nodes represent bus stops, edges represent bus routes, and a label on each edge represents the club cheered on by the bus route driver.

There is a key observation for this problem: if two nodes $$$\mathbf{P_j}$$$ and $$$\mathbf{C_j}$$$ are connected, we would be able to traverse all the edges in the connected component in the path from $$$\mathbf{P_j}$$$ to $$$\mathbf{C_j}$$$. The proof for this observation is provided at the end of this analysis. With this observation, we can find that if the total number of distinct edge clubs in this connected component is odd, there is always a path from $$$\mathbf{P_j}$$$ to $$$\mathbf{C_j}$$$ that can walk on an odd number of clubs by traversing all the edges in the connected component at least once.

Therefore, for each query, we can check if there exists a subgraph that satisfies:

• $$$\mathbf{P_j}$$$ and $$$\mathbf{C_j}$$$ are connected.
• There is an odd number of distinct edge clubs in the connected component that contains $$$\mathbf{P_j}$$$ and $$$\mathbf{C_j}$$$.

Test Set 1

Since the number of distinct clubs $$$\mathbf{K}$$$ is small in Test Set 1, we can enumerate all combinations of an odd number of distinct clubs. For each combination, build a new graph $$$G'$$$ by removing edges with clubs not in the combination. Afterwards, for each query we can check if $$$\mathbf{P_j}$$$and $$$\mathbf{C_j}$$$ are connected in any $$$G'$$$, and if there is an odd number of edge clubs in the corresponding connected component by iterating through all the edges in the connected component.

We can use Disjoint Set Union (DSU) to find the connected componnets in $$$G'$$$ and all the nodes and edges in them. A connected component has edge club $$$c$$$ if and only if $$$u$$$ and $$$v$$$ are in this component (in the same set in DSU) and the edge $$$(u, v)$$$ has club $$$c$$$. We can prebuild the DSU for all $$$G'$$$s and find the number of distinct clubs for each of the connected component in them. For each query, we can iterate through all prebuilt DSUs and check if $$$\mathbf{P_j}$$$ and $$$\mathbf{C_j}$$$ are connected in a component with odd number of edge clubs in it.

Space complexity:

There are at most $$$2^\mathbf{K}$$$ $$$G'$$$s, and each graph has at most $$$\mathbf{N}$$$ connected components:

• Disjoint Set Unions(DSUs): $$$O(2^\mathbf{K} \times \mathbf{N})$$$
• Number of distinct clubs in connected components in graphs: $$$O(2^\mathbf{K} \times \mathbf{N})$$$

Time complexity:

• Build DSUs and component clubs set (number of distinct clubs): $$$O(2^\mathbf{K} \times (\mathbf{N} + \mathbf{M}))$$$
• Per query: $$$O(2^\mathbf{K})$$$ for checking all DSUs and number of distinct clubs in the corresponding connected component.
• Overall: $$$O(2^\mathbf{K} \times (\mathbf{N} + \mathbf{M}) + \mathbf{Q} \times 2^\mathbf{K}) = O(2^\mathbf{K} \times (\mathbf{N} + \mathbf{M} + \mathbf{Q}))$$$

Test Set 2 and Test Set 3

We can improve our solution by reducing the club combinations to check: we just need to check the original graph $$$G$$$ and all $$$G'$$$ built by removing edges with one club. We can prove this statement by considering all possible cases for query $$$\mathbf{P_j}$$$ and $$$\mathbf{C_j}$$$:

Let $$$\text{Clubs}(G, \mathbf{P_j}, \mathbf{C_j})$$$ be the set of distinct clubs in the connected component containing $$$\mathbf{P_j}$$$ and $$$\mathbf{C_j}$$$ in graph $$$G$$$, and $$$|\text{Clubs}(G, \mathbf{P_j}, \mathbf{C_j})|$$$ be the number of distinct clubs:

• If $$$\mathbf{P_j}$$$ and $$$\mathbf{C_j}$$$ are not connected in $$$G$$$, there is no valid path for this query.
• If $$$\text{Clubs}(G, \mathbf{P_j}, \mathbf{C_j})$$$ has an odd number of distinct clubs ($$$|\text{Clubs}(G, \mathbf{P_j}, \mathbf{C_j})| = 2n + 1$$$), there is always a valid path walking on an odd number of distinct clubs.
• Otherwise, if there is a path from $$$\mathbf{P_j}$$$ to $$$\mathbf{C_j}$$$ composed of edges with odd number ($$$2n + 1$$$) of distinct clubs, $$$\mathbf{P_j}$$$ and $$$\mathbf{C_j}$$$ will always be connected after removing edges with those $$$\mathbf{K} - (2n + 1)$$$ clubs not in that path from graph $$$G$$$. Then, let $$$G_{minimal}$$$ be graph after removing those edges, we can find another path with an odd number of clubs by traversing additional connected edges with two extra clubs $$$c_1$$$ and $$$c_2$$$ where $$$c_1, c_2 \notin \text{Clubs}(G_{minimal}, \mathbf{P_j}, \mathbf{C_j})$$$ and $$$c_1, c_2 \in \text{Clubs}(G, \mathbf{P_j}, \mathbf{C_j})$$$. Recursively, we can find a valid path in $$$G'$$$ by removing only one club from $$$\text{Clubs}(G, \mathbf{P_j}, \mathbf{C_j})$$$. It means if there is a valid path, there is always a $$$G'$$$ where $$$|\text{Clubs}(G', \mathbf{P_j}, \mathbf{C_j})|$$$ is odd and built by removing edges of one club.

Therefore, we just need to build the DSUs and connected components clubs sets for the $$$\mathbf{K}+1$$$ graphs that result from removing $$$0$$$ or $$$1$$$ club from the original graph.

Space complexity:

There are $$$\mathbf{K} + 1$$$ $$$G'$$$s, and each graph has at most $$$\mathbf{N}$$$ connected components:

• Disjoint Set Unions(DSUs): $$$O((\mathbf{K} + 1) \times \mathbf{N}) = O(\mathbf{K} \times \mathbf{N})$$$
• Number of distinct clubs in connected components in graphs: $$$O(\mathbf{K} \times \mathbf{N})$$$

Time complexity:

• Build DSUs and component clubs set (number of distinct clubs): $$$O((\mathbf{K} + 1) \times (\mathbf{N} + \mathbf{M})) = O(\mathbf{K} \times (\mathbf{N} + \mathbf{M}))$$$
• Per query: $$$O(\mathbf{K})$$$ for checking all DSUs and number of distinct clubs in the corresponding connected component.
• Overall: $$$O(\mathbf{K} \times (\mathbf{N} + \mathbf{M} + \mathbf{Q}))$$$

Test Set 2 does not require prebuilding and sharing the DSU and clubs sets across queries. Solutions with time complexity $$$O(\mathbf{Q} \times (\mathbf{K} \times (\mathbf{N} + \mathbf{M})))$$$ that build DSUs and clubs sets for each of the query separately are acceptable.

Proof: Path traversing all edges in the connect component always exists

Suppose there is a path from $$$v_1$$$ to $$$v_n$$$ where $$$p=(v_1,v_2,\dots,v_{n-1}, v_n)$$$, and let $$$\text{Nodes}(p) = \{v_1, v_2, \dots, v_n\}$$$ to be the set of nodes visited at least once in path $$$p$$$:

1. Node Reachability: For any node $$$v_i \in \text{Nodes}(p)$$$, if there is a node $$$u_m$$$ and a path from $$$v_i$$$ to $$$u_m$$$ where $$$p'=(v_i, u_1, u_2, \dots, u_m)$$$ in the graph, there is always a path $$$p_{merged}$$$ from $$$v_1$$$ to $$$v_n$$$ where $$$\text{Nodes}(p_{merged}) = \text{Nodes}(p) \cup \text{Nodes}(p') \supseteq \text{Nodes}(p) \cup \{u_m\}$$$. A valid path would be $$$p_{merged}=(v_1, \dots,v_i, u_1,\dots, u_{m-1}, u_m, u_{m-1}, \dots, u_1, v_i, v_{i+1}, \dots, v_n)$$$. Notice that we can visit same node or edge multiple times.
2. Edge Reachability: For any node $$$v_i \in \text{Nodes}(p)$$$, and for edges $$$(v_i, u_1), (v_i, u_2), \dots, (v_i, u_m)$$$ directly connect to $$$v_i$$$, there always exists a path $$$p'_{merged}$$$ from $$$v_1$$$ to $$$v_n$$$ which traverses all these edges at least once. A valid path would be $$$p'_{merged}=(v_1, \dots,v_i, u_1, v_i, u_2, \dots, v_i, u_m, v_i, v_{i+1}, \dots, v_n)$$$.
3. From (1), we can derive that if nodes $$$\mathbf{P}$$$ and $$$\mathbf{C}$$$ are connected, there exist a path from $$$\mathbf{P}$$$ to $$$\mathbf{C}$$$ which visits all reachable nodes from $$$\mathbf{P}$$$ at least once, which are all nodes in the connected component.
4. From (2) and (3), we can derive that if nodes $$$\mathbf{P}$$$ and $$$\mathbf{C}$$$ are connected, there exists a path which traverses all the edges in the connected component at least once.

Let $$$N$$$ be the length of $$$\mathbf{S}$$$.

Test Set 1

In Test Set 1, we have either $$$\mathbf{P}=2$$$ or $$$\mathbf{P}=3$$$. We can solve each of those cases separately:

1. If $$$\mathbf{P} = 2$$$, there are exactly $$$N-1$$$ possible places to split $$$\mathbf{S}$$$ into two parts. We can try them all, and for each one, check whether Badari can win by using a solution to problem Game Sort: Part 1. Then we know that Amir wins if any of those $$$N-1$$$ places results in a situation where Badari cannot win.
2. If $$$\mathbf{P} = 3$$$, there are exactly $$$\frac{(N-1)(N-2)}{2}$$$ possible ways to separate: as before, there are $$$N-1$$$ possible places to split, and we must now choose two of those in order to separate $$$\mathbf{S}$$$ into three parts. Just like before, one option is to try all such pairs, and for each one, check whether Badari can win.

Such an algorithm has a time complexity $$$O(N^3 \lg N)$$$ or perhaps $$$O(N^3)$$$ depending on implementation details, so it is enough to pass Test Set 1.

Test Set 2

First of all, we should notice that, whenever $$$\mathbf{P} = N$$$, Amir has no choice and $$$\mathbf{S}$$$ has to be separated into its $$$N$$$ individual letters. Furthermore, for each single-letter part Badari has no choice of rearrangement. Thus, Badari wins in this case if and only if the letters of $$$\mathbf{S}$$$ are already sorted in non-decreasing order.

This is a special case to check separately, as many of the properties that we will use later to solve Test Set 2 are false if $$$\mathbf{P} = N$$$. From now on, we assume $$$\mathbf{P} \lt N$$$.

We can solve the case $$$\mathbf{P} = 2$$$ as in Test Set 1, trying all $$$N-1$$$ possible ways to split $$$\mathbf{S}$$$. To avoid doing so in quadratic time, we can keep two arrays with a count of each character per part. Then, when moving from position $$$i$$$ to $$$i+1$$$, we only need to increment one count and decrement the other, and use the check from the solution for Game Sort: Part 1. We thus get an $$$O(N)$$$ solution for $$$\mathbf{P}=2$$$.

Finally, the only remaining case is $$$3 \leq \mathbf{P} \lt N$$$. Having at least three parts allows Amir to win lots of strings easily. For instance, for any string which does not start with its alphabetically smallest character (let us call it $$$c$$$), Amir can win by simply splitting the first occurrence of $$$c$$$ as a single-letter part.

Similarly, if the alphabetically smallest character $$$c$$$ appears anywhere on the string that is not the first or second character, it is possible to win by splitting that occurrence of $$$c$$$ as a single-letter part, while leaving the first and second characters of the string unsplit (it is possible because $$$\mathbf{P} \lt N$$$).

In the remaining case for $$$\mathbf{P} \geq 3$$$, $$$\mathbf{S}$$$ has the following property:

• $$$c$$$ appears in $$$\mathbf{S}$$$ either exactly once as the first character of $$$\mathbf{S}$$$, or exactly twice as the first two characters of $$$\mathbf{S}$$$.

Let us call a string having that property a bad string, and any string without the property good. The only remaining case to solve is for bad strings and $$$\mathbf{P} \geq 3$$$.

We can assume that Badari plays optimally, and that she follows the optimal strategy explained in the solution of Game Sort: Part 1. That strategy implies that, for a bad string, no matter how we separate $$$\mathbf{S}$$$ into parts, after Badari's rearrangement, the first part is lexicographically lower than or equal to all the other parts. This is because the first part will either contain a single $$$c$$$, or every occurrence of $$$c$$$.

Thus, to split a bad string into $$$\mathbf{P}$$$ parts, we must choose the length $$$l$$$ of the first part, and then separate the remaining $$$N-l$$$ characters into $$$\mathbf{P}-1$$$ parts. Note that the first part will be in sorted order, and by the property of bad strings, it will be the lexicographically smallest part and will not impose restrictions on the rearrangement of the second part. So Amir wins such a split if and only if he wins the split of the last $$$N-l$$$ characters into $$$\mathbf{P}-1$$$ parts.

Using all these ideas, we can recursively compute, for each suffix of the initial string $$$\mathbf{S}$$$ and each $$$k$$$ with $$$2 \leq k \leq \mathbf{P}$$$, who wins when the game is played for that suffix as initial string and $$$k$$$ as the number of parts.

Note that, with this approach, computing who wins for each suffix when splitting into $$$k=2$$$ parts must be done independently as a base case. It is possible to compute all such values efficiently, but it involves even more case analysis and careful implementation, so we will leave it as an exercise for the reader. Such an approach would lead to an overall $$$O(N \times \mathbf{P})$$$ time solution, which should pass if implemented carefully.

Although the main idea of this solution is relatively direct, the implementation has many special cases and details, and is tricky to code quickly and correctly. This approach can be implemented more efficiently and in a simpler way, by not storing all of the $$$N \times \mathbf{P}$$$ subproblems. Instead, we can store, for each $$$k$$$ between $$$2$$$ and $$$\mathbf{P}$$$, the index of the shortest suffix that Amir wins when splitting that suffix into $$$k$$$ parts. In a sense, the shortest suffix that works is the only relevant working suffix, because when choosing the length $$$l$$$ of a first part to cut, if any winning suffix is available, then the shortest one is. This key observation also avoids having to code the general problem of identifying all winning suffixes for $$$k=2$$$, since we only need to find the shortest such suffix now, which is quite a simpler problem that we will also leave as an exercise. (Hint: focus only on which characters are lower, equal to, or higher than the very last character of the input string.) With this, we get an $$$O(N)$$$ solution.

Another approach

There is another approach which is quite simpler to code, but maybe harder to find and prove correct. We will sketch the algorithm and leave the proof of correctness as an exercise for the reader.

We solve cases $$$\mathbf{P}=N$$$ and $$$\mathbf{P}=2$$$ exactly as before. Then we observe that, for $$$\mathbf{P} \geq 4$$$, Amir has even more favorable strings:

1. If $$$\mathbf{S}$$$ is not sorted, then Amir wins by taking the first consecutive pair of letters in $$$\mathbf{S}$$$ that are out of order, and separating them both into single-letter parts.
2. If $$$\mathbf{S}$$$ contains three consecutive identical letters, then Amir wins by splitting so that the first two occurrences form a part of length $$$2$$$, and the third consecutive occurrence forms a single-letter part.
3. If neither (1) nor (2) hold, then it is easy to see that Amir cannot win.

Only the hardest case where $$$\mathbf{P}=3$$$ remains. Suppose that Amir splits $$$\mathbf{S}$$$ into three parts, say $$$A,B,C$$$ (so $$$\mathbf{S}=ABC$$$). It is clear that, if partitioning string $$$AB$$$ into $$$A,B$$$ is a win for Amir with $$$\mathbf{P}=2$$$ and input string $$$AB$$$, then $$$A,B,C$$$ works for $$$\mathbf{P}=3$$$ and input string $$$\mathbf{S}=ABC$$$. Similarly, if partitioning $$$BC$$$ into $$$B,C$$$ is a win for Amir for $$$\mathbf{P}=2$$$ and input string $$$BC$$$, then also $$$A,B,C$$$ for $$$\mathbf{P}=3$$$ and input $$$\mathbf{S}=ABC$$$.

The surprising result is that, for any string $$$\mathbf{S}$$$, Amir wins if and only if one of these two cases occur. This is not obvious: There are examples of partitions $$$A,B,C$$$ that work, but neither $$$A,B$$$ nor $$$B,C$$$ work; e.g. XY,AZ,XY for input string XYAZXY. However, in any such case there will be some other partition that works and has this property, e.g. XY,A,ZXY for input string XYAZXY, as XY,A works.

Finally, we only need to check if Amir wins any prefix or suffix of $$$\mathbf{S}$$$ with $$$\mathbf{P}=2$$$. Note that this is actually simpler than computing for every suffix and prefix whether Amir wins with $$$\mathbf{P}=2$$$, and can be done by a slight modification of the full solution to the $$$\mathbf{P}=2$$$ case, which we leave as an exercise.