Solution Explanation: Minimum Time to Kill All Monsters

This problem involves finding the minimum number of days required to defeat all monsters, given their individual powers and a daily mana gain that increases with each monster defeated. The key to solving this efficiently is recognizing that the number of monsters is small enough (up to 17) to allow for a bit manipulation approach using bitmasking.

Core Idea:

We represent the state of the monsters using a bitmask. Each bit in the mask corresponds to a monster:

• 0: Monster is alive.
• 1: Monster is defeated.

The algorithm explores all possible sequences of monster defeats by iterating through all possible bitmasks. This is a form of state-space search, optimized using dynamic programming (or memoization, which achieves the same effect).

Algorithms:

Two approaches are presented: memoization (top-down) and dynamic programming (bottom-up). Both have the same time and space complexity.

1. Memoization (Top-Down):

• dfs(mask) function: This recursive function takes a bitmask (mask) as input, representing the current state of monsters. It returns the minimum number of days needed to defeat all remaining monsters in that state.
• Base Case: If mask is 0 (all monsters defeated), it returns 0 days.
• Recursive Step: It iterates through all live monsters (bits set to 0). For each monster, it calculates the number of days required to defeat that monster ((monsterPower + gain - 1) // gain), adds it to the minimum days needed to defeat the remaining monsters (dfs(mask ^ (1 << monsterIndex))), and updates the minimum days found so far.
• Caching: The results of dfs(mask) are stored in a cache (using Python's @cache decorator or similar mechanisms in other languages) to avoid redundant computations.

2. Dynamic Programming (Bottom-Up):

• f[mask] array: An array where f[mask] stores the minimum number of days to defeat monsters in state mask. It's initialized with infinity for all masks except f[0] = 0.
• Iteration: The algorithm iterates through all possible bitmasks from 1 to (1 << n) - 1, where n is the number of monsters.
• Transition: For each mask, it iterates through defeated monsters (bits set to 1) and calculates the minimum days based on the previous state (f[mask ^ (1 << monsterIndex)]) and the cost of defeating the current monster.

Time Complexity: O(2n * n)

The algorithm iterates through all 2n possible bitmasks (states), and for each mask, it iterates through at most n monsters. This leads to an exponential time complexity, which is unavoidable for this type of exhaustive search. However, because n is limited to 17, this approach remains practical.

Space Complexity: O(2n)

The space complexity is dominated by the cache (memoization) or the DP array, which stores results for all 2n possible bitmasks.

Code Examples:

(See the comprehensive code examples provided in the original response, including Python, Java, C++, Go, and TypeScript. The code implementations directly reflect the algorithms described above.)

In summary: This problem demonstrates the effective use of bit manipulation and dynamic programming (or memoization) to solve a combinatorial optimization problem. The key is to efficiently represent the state of the problem using bitmasks and avoid redundant calculations through caching or dynamic programming. While the time complexity is exponential, the constraint on the number of monsters makes the solution feasible.