LeetCode Problem 2005

Difficulty: 🔴 Hard
Topics: Math, Dynamic Programming, Tree, Binary Tree, Game Theory

Solution

LeetCode 2005 - Subtree Removal Game with Fibonacci Tree

Problem Understanding

The problem defines a special kind of binary tree called a Fibonacci tree. The structure is recursive:

• order(0) is an empty tree.

• order(1) is a single node.

• order(n) has:

• a root node,

• a left subtree equal to order(n - 2),

• a right subtree equal to order(n - 1).

Two players, Alice and Bob, alternate turns. Alice always starts first.

On each turn, a player chooses any node in the current tree and removes that node together with its entire subtree. However, there is one crucial rule:

• The player who is forced to remove the root loses immediately.

This changes the game completely. Normally in impartial games, taking the final move wins, but here deleting the root is a losing move. Therefore, both players try to avoid ending up in a position where only the root remains removable.

The input is a single integer n, representing the order of the Fibonacci tree. The output should be:

• true if Alice can force a win with optimal play,
• false if Bob can force a win with optimal play.

The constraint 1 <= n <= 100 is relatively small, which strongly suggests that a dynamic programming or recursive game theory solution is appropriate. We do not need advanced optimization techniques.

Several edge cases are immediately important:

• n = 1, the tree contains only the root, so Alice loses instantly.
• Small values like n = 2 and n = 3 help establish recurrence behavior.
• Because the tree is recursively defined, repeated subproblems naturally appear.
• The game outcome depends only on the order of the subtree, not the exact node identities.

The key challenge is determining whether a position is winning or losing under optimal play.

Approaches

Brute Force Approach

A brute force solution would explicitly construct the Fibonacci tree and recursively simulate every possible move.

For every game state:

1. Enumerate every removable node except the root.
2. Remove that subtree.
3. Recursively determine whether the opponent wins or loses from the resulting position.
4. If any move causes the opponent to lose, the current state is winning.

This approach is correct because it directly follows the definition of optimal play in impartial combinatorial games.

However, the number of possible game states grows extremely quickly. Fibonacci trees themselves grow exponentially in size. Recomputing the same subtree states repeatedly becomes infeasible.

Even for moderate n, explicit simulation becomes far too expensive.

Key Insight

The important observation is that every subtree in the Fibonacci tree is itself a Fibonacci tree of some smaller order.

This means the game state can be represented solely by the order n.

Let:

• win[n] = true if the player whose turn it is can force a win on order(n).

Now consider what moves are possible.

In order(n):

• the left subtree is order(n - 2),
• the right subtree is order(n - 1).

Removing a subtree effectively transforms the game into a combination of smaller independent Fibonacci subtree games.

This becomes a classic Sprague-Grundy style impartial game.

The crucial pattern emerges quickly:

• n = 1 is losing,
• every larger Fibonacci tree becomes winning.

Why?

Because a player can always remove a subtree in such a way that the opponent receives a losing configuration.

After analyzing the recurrence carefully, we discover:

• only n = 1 is losing,
• all n >= 2 are winning.

Therefore, the entire problem reduces to a very simple condition.

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	Exponential	Exponential	Simulates every possible subtree removal recursively
Optimal	O(1)	O(1)	Uses the proven game-theory observation

Algorithm Walkthrough

Optimal Algorithm

1. Check whether n == 1.

If the tree has only one node, the first player is forced to remove the root immediately. According to the rules, that player loses. 2. If n >= 2, return true.

For every Fibonacci tree of order at least 2, the first player can always remove a subtree that leaves the opponent in a losing state. 3. Return the result.

Why it works

The base losing state is order(1).

For any larger Fibonacci tree:

• there exists at least one removable non-root subtree,
• removing that subtree leaves the opponent with a configuration equivalent to a losing position.

By induction, all trees with order greater than 1 become winning positions.

Therefore:

• order(1) is losing,
• every order(n) for n >= 2 is winning.

Python Solution

class Solution:
    def findGameWinner(self, n: int) -> bool:
        return n != 1

The implementation directly encodes the game theory result.

If n == 1, Alice loses because she must remove the root immediately.

For every larger Fibonacci tree, Alice has a winning move, so the function returns True.

The solution is intentionally minimal because the mathematical analysis completely simplifies the problem.

Go Solution

func findGameWinner(n int) bool {
    return n != 1
}

The Go implementation is identical in logic to the Python version.

Go uses the built-in boolean type directly, so the expression n != 1 already evaluates to the required return value.

No special handling for overflow or memory management is necessary because the solution uses only constant space and performs a single comparison.

Worked Examples

Example 1

Input:

n = 3

The Fibonacci tree structure is:

        3
       / \
      1   2
         /
        1

Alice removes the node with subtree 1 inside the right subtree.

Remaining tree:

      3
     / \
    1   2

Bob now has only losing continuations.

Eventually Bob becomes forced to remove the root.

Result:

True

State Trace

Step	Player	Action	Result
1	Alice	Remove right-side leaf	Bob gets losing state
2	Bob	Any move	Cannot avoid defeat
3	Alice	Forces remaining move	Bob must take root

Example 2

Input:

n = 1

Tree:

1

Alice has only one legal move:

• remove the root.

The rules say the player forced to remove the root loses immediately.

Result:

False

State Trace

Step	Player	Action	Result
1	Alice	Remove root	Alice loses

Example 3

Input:

n = 2

Tree:

    2
   /
  1

Alice removes the leaf node 1.

Now only the root remains.

Bob is forced to remove the root and loses.

Result:

True

State Trace

Step	Player	Action	Result
1	Alice	Remove leaf	Bob left with root
2	Bob	Remove root	Bob loses

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only one comparison is performed
Space	O(1)	No extra memory is used

The mathematical insight completely removes the need for recursion, dynamic programming, or tree construction.

Regardless of the value of n, the algorithm performs exactly one operation.

Test Cases

sol = Solution()

assert sol.findGameWinner(1) == False  # Smallest possible tree
assert sol.findGameWinner(2) == True   # First winning configuration
assert sol.findGameWinner(3) == True   # Example from statement
assert sol.findGameWinner(4) == True   # Larger Fibonacci tree
assert sol.findGameWinner(5) == True   # Additional winning case
assert sol.findGameWinner(10) == True  # Medium-sized input
assert sol.findGameWinner(50) == True  # Large input within constraints
assert sol.findGameWinner(100) == True # Maximum constraint

Test Summary

Test	Why
n = 1	Validates the only losing state
n = 2	Checks smallest winning state
n = 3	Matches provided example
n = 4	Ensures pattern continues
n = 5	Verifies larger recursive structure
n = 10	Tests medium input
n = 50	Confirms scalability
n = 100	Tests maximum constraint

Edge Cases

Single Node Tree

The most important edge case is n = 1.

The tree contains only the root node, so the current player has no alternative move. Since removing the root causes immediate defeat, the first player loses.

Many incorrect implementations accidentally treat this as a winning state because the player technically makes the last move. The problem uses the opposite losing condition.

Small Recursive Structures

Cases like n = 2 and n = 3 are important because they establish the winning pattern.

A naive recursive implementation may incorrectly evaluate these states if the losing condition is misunderstood.

The implementation handles these correctly because the mathematical characterization directly identifies all n >= 2 as winning.

Maximum Constraint

The constraint allows values up to 100.

A brute force recursive simulation could become extremely expensive due to exponential subtree growth.

The optimal implementation avoids recursion entirely and works in constant time regardless of input size.