Problem Description

Alice and Bob are engaged in a fantasy battle game, which unfolds over n rounds. In each round, they summon one of three magical creatures: a Fire Dragon, a Water Serpent, or an Earth Golem. The creatures have specific interactions with each other:

• A Fire Dragon beats an Earth Golem.
• A Water Serpent beats a Fire Dragon.
• An Earth Golem beats a Water Serpent.
• If both players summon the same creature, it's a tie with no points awarded.

Given Alice's sequence of moves as a string s with characters 'F' for Fire Dragon, 'W' for Water Serpent, and 'E' for Earth Golem, Bob needs to choose his moves. It's guaranteed that Bob won't summon the same creature in consecutive rounds. Bob wins if his total points exceed Alice's after all rounds. The task is to determine how many sequences Bob can use to beat Alice, with the final answer modulo (10^9 + 7).

Intuition

To solve the problem, we utilize a recursive approach with memoization. The function dfs(i, j, k) is defined as follows:

• i is the index in string s where the current round starts.
• j is the score difference between Bob and Alice.
• k is the last summoned creature by Bob.

The goal is to determine how many sequences of moves Bob can make that results in a win over Alice. We begin the exploration with dfs(0, 0, -1), where -1 signifies Bob hasn't summoned any creatures yet.

The recursive strategy involves:

• Checking if remaining rounds (n - i) are inadequate for altering the score difference such that it's in Bob's favor (i.e., j being positive).
• Ending recursion when all rounds are complete (i >= n), returning 1 if Bob's score is greater, otherwise returning 0.
• Iterating over possible creatures Bob can summon, ensuring no repetition of creatures in consecutive rounds, and employing the helper function calc(x, y) to update the score based on outcomes of the summoned creatures.

This solution efficiently maps characters to integers and calculates possible winning sequences using a depth-first search, systematically summing them while applying modulo (10^9 + 7) to handle large numbers.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution uses a memoization search strategy, combining recursion and caching to explore all potential sequences of Bob's moves efficiently. Here's a breakdown of the implementation:

1. DFS Function Definition: We define a function dfs(i, j, k) to recursively determine the number of winning sequences Bob can employ:

   • i represents the current index in Alice's move sequence s.
   • j is the score difference between Bob and Alice (positive if Bob leads).
   • k is the last creature Bob summoned.

2. Base Conditions:

   • If the remaining rounds aren't sufficient for Bob to overtake Alice (n - i <= j), return 0.
   • If all rounds are concluded (i >= n), check if Bob's score leads (j < 0). Return 1 if true, otherwise 0.

3. Recursion with Choices:

   • Each recursion step considers all three creatures Bob can summon (0, 1, 2, corresponding to "F", "W", "E"), skipping the previous creature to ensure no consecutive repeats.
   • Use the helper function calc(x, y) to compute points based on the interaction of Alice's and Bob's choices.

4. Score Calculation:

   • The helper calc(x, y) returns the point effect based on Bob's and Alice's summoned creatures. Specifically:
     • Win situations: (0, 2), (1, 0), (2, 1) return 1.
     • Loss situations are the reverse of win cases.
     • Ties return 0.

5. Caching for Optimization:

   • Use @cache to memoize results of dfs(i, j, k), preventing redundant calculations and improving efficiency.

6. Modulo Operation:

   • Since potential outcomes can be extensive, results are taken modulo (10^9 + 7).

This solution effectively evaluates all possible arrangements of Bob's moves, using recursive exploration with memoization to achieve the desired result within computational limits.

Example Walkthrough

Let's consider a small example to illustrate the solution approach. Assume Alice's sequence of moves is "FWE", which represents 3 rounds where Alice plays:

1. Round 1: Fire Dragon ('F')
2. Round 2: Water Serpent ('W')
3. Round 3: Earth Golem ('E')

The objective for Bob is to choose his moves in such a way that he wins more rounds than Alice, while ensuring no creature is summoned consecutively.

Step-by-Step Process:

1. Initialize Variables:

   • Start with dfs(0, 0, -1), indicating we begin at the first round, with a score difference of zero, and no prior creature summoned by Bob.

2. Round 1 Choices:

   • Alice plays 'F'.
   • Bob considers three possibilities: 'W', 'E', 'F'. Based on the rule that Bob can't repeat creatures, we note:
     • If Bob chooses 'W': Bob beats Alice, so the score difference becomes +1 (Bob is leading).
     • If Bob chooses 'E': Bob is defeated by Alice, the score difference becomes -1 (Alice is leading).
     • If Bob chooses 'F': It's a tie, score difference remains 0.

3. Round 2 Choices:

   • Move on to dfs(1, score_difference, last_choice).
   • Alice plays 'W'.
   • If Bob's previous choice was 'W', he must choose between 'F' or 'E':
     • 'E' beats 'W': Bob gains a point, score difference increases by 1.
     • 'F' loses to 'W': Bob loses a point, score decreases by 1.
   • If Bob’s previous choice was 'E' or 'F', adjust accordingly while ensuring no repeat of the immediate prior choice.

4. Round 3 Choices:

   • Process dfs(2, score_difference, last_choice).
   • Alice plays 'E'.
   • Again, make sure not to repeat Bob’s Round 2 choice:
     • If Bob chooses 'F', Bob wins the round.
     • If Bob chooses 'W', Bob loses the round.
     • Previous choice-aware adjustments continue here.

5. Evaluate Final Outcome:

   • At the end of all rounds (i = n), if Bob's score is greater than Alice’s (score_difference > 0), count this sequence as a valid win for Bob.
   • Sum all valid sequences modulo (10^9 + 7) to handle large results.

Through recursive exploration, we evaluate each sequence of Bob’s moves, tracking scores and respecting non-repetition rules, using memoization to enhance computation efficiency. This walkthrough demonstrates the approach to finding all possible winning move sequences for Bob given Alice's predefined moves.

Solution Implementation

Time and Space Complexity

The function dfs is called recursively with three parameters: i, j, and k, where i is the current index, j represents a cumulative score, and k is the last move. The recursion explores all possible sequences of moves.

Time Complexity

The function's complexity arises from the potential number of calls made due to different combinations of the parameters. The potential states are determined by:

• The length of the string s, denoted as n.
• The range of cumulative scores j, which depends on the length and can go up to O(n).
• The possible moves represented by k, which can be one of 3 values: 0, 1, or 2.

Thus, the number of possible states is O(n) * O(n) * O(k), where k is the size of the character set (3 different moves). Therefore, the time complexity is O(n^2 * k).

Space Complexity

The space complexity is due to the cache used for memoization and the recursion stack:

• The memoization cache stores results for combinations of (i, j, k). This results in a space requirement of O(n^2 * k).
• The recursion depth is O(n), contributing an additional space factor.

Thus, the overall space complexity is O(n^2 * k).

Learn more about how to find time and space complexity quickly.