Chalkboard XOR Game

1. Define a recursive function, say canWin(currentNums, memo), that returns true if the current player can win from the state defined by currentNums.
2. The state can be uniquely identified by the list of numbers. To use it as a key in a memoization map, we can sort the list and convert it to a string or use the list object itself (if its hashCode and equals methods are handled correctly).
3. Base Case/Winning Condition: Calculate the bitwise XOR sum X of all numbers in currentNums. If X == 0, the current player wins immediately. Return true.
4. Recursive Step: The current player needs to find a move that leads to a state from which the opponent cannot win.
5. Iterate through each number num in currentNums.
6. A move consists of removing num. This move is a losing move for the current player if the new XOR sum becomes 0. The new XOR sum is X ^ num. So, the player must choose a num such that X ^ num != 0.
7. For each such valid move, create the next state nextNums = currentNums - {num}.
8. Recursively call canWin(nextNums, memo). If this call returns false, it means we found a move where the opponent loses. Therefore, the current player wins. We can return true.
9. If after trying all valid moves, none lead to a state where the opponent loses (i.e., all recursive calls return true), then the current player has no winning move and will lose. Return false.
10. Before returning, store the result in the memoization table to avoid recomputing for the same state.

This approach simulates the game turn by turn. We define a function canWin which takes the current list of numbers on the chalkboard. To avoid re-calculating the result for the same set of numbers, we use a map for memoization, where the key represents the state (the list of numbers) and the value is the boolean result.

The logic inside canWin is as follows:

• First, check if the result for the current state is already in our memoization table. If so, return the cached value.
• Calculate the XOR sum of the current numbers. If it's 0, the current player wins by the game's rules. We store true and return it.
• If the XOR sum is not 0, the player must make a move. A move is picking a number num to erase. The player loses if this move makes the new XOR sum 0. So, a player must pick a num such that current_xor_sum ^ num != 0.
• We iterate through all possible valid moves. For each move, we check if the opponent can win from the resulting state by making a recursive call. If we find any move for which the opponent cannot win (the recursive call returns false), it means we have found a winning line of play. We can then cache and return true.
• If we iterate through all valid moves and find that the opponent always has a winning response, then the current player will lose from this state. We cache and return false.

1import java.util.*;2 3class Solution {4    public boolean xorGame(int[] nums) {5        List<Integer> initialNums = new ArrayList<>();6        for (int num : nums) {7            initialNums.add(num);8        }9        // Sorting is important for consistent map keys if lists are used directly10        // A better key would be a canonical string representation.11        Collections.sort(initialNums);12        return canWin(initialNums, new HashMap<>());13    }14 15    private boolean canWin(List<Integer> nums, Map<List<Integer>, Boolean> memo) {16        if (nums.isEmpty()) {17            return false; // XOR sum is 0, previous player lost, so current player wins.18                          // This case is handled by the xorSum == 0 check.19        }20 21        if (memo.containsKey(nums)) {22            return memo.get(nums);23        }24 25        int xorSum = 0;26        for (int num : nums) {27            xorSum ^= num;28        }29 30        if (xorSum == 0) {31            memo.put(nums, true);32            return true;33        }34 35        // Try all possible moves36        for (int i = 0; i < nums.size(); i++) {37            int numToRemove = nums.get(i);38            39            // A player loses if their move makes the XOR sum 0.40            // So, a player must make a move that does NOT result in 0.41            if ((xorSum ^ numToRemove) != 0) {42                List<Integer> nextNums = new ArrayList<>(nums);43                nextNums.remove(i);44                45                // If the opponent cannot win from the next state, we win.46                if (!canWin(nextNums, memo)) {47                    memo.put(nums, true);48                    return true;49                }50            }51        }52 53        // If all moves lead to a state where the opponent wins, we lose.54        memo.put(nums, false);55        return false;56    }57}

Note: The above code is conceptually correct but will be too slow for the given constraints. Using a sorted list as a map key is also inefficient.

1. Calculate the bitwise XOR sum of all elements in the nums array.
2. Get the total number of elements, N, from the length of the array.
3. Apply the winning condition logic:
   • If the initial XOR sum is 0, Alice wins immediately on her first turn.
   • If the initial XOR sum is not 0, the outcome depends on the number of elements N.
   • If N is even, Alice is guaranteed to win.
   • If N is odd, Alice will lose (assuming optimal play from Bob).
4. Return true if xor_sum == 0 or N % 2 == 0, and false otherwise.

Let's analyze the game rules to find a winning strategy.

Winning/Losing Conditions:

1. A player wins if, at the start of their turn, the XOR sum of all numbers on the board is 0.
2. A player loses if they make a move (erase a number num) that causes the XOR sum of the remaining numbers to become 0.

Let X be the XOR sum of the numbers on the board at the start of a player's turn.

Case 1: Initial XOR sum is 0. If the XOR sum of the initial nums array is 0, Alice starts her turn and wins immediately. So, if X_initial == 0, Alice wins.

Case 2: Initial XOR sum is not 0. Alice must make a move. Let the current XOR sum be X and the number of elements be n. Alice must choose a number num to erase. The new XOR sum will be X ^ num. To not lose, she must choose num such that X ^ num != 0, which means num != X.

Can a player always make a non-losing move? A player is forced to lose if, for every number num_i on the board, num_i = X. Let's see when this can happen. If all n numbers on the board are equal to X, their total XOR sum is X ^ X ^ ... ^ X (n times).

• If n is even, this sum is 0. This contradicts our assumption that X != 0.
• If n is odd, this sum is X. This is consistent.

This means a player can only be forced into a losing position if they face a board with an odd number of elements. If a player faces a board with an even number of elements and a non-zero XOR sum, it's impossible for all numbers to be equal to the XOR sum. Thus, there must be at least one number num such that num != X. The player can safely remove this num and pass the turn.

Applying this to Alice and Bob:

• If nums.length is even: Alice starts with an even number of elements. Since the initial XOR sum is not 0, she can always make a non-losing move. After her move, the board will have an odd number of elements for Bob. Bob will then make a move, leaving an even number for Alice. This continues. Alice always plays on a board with an even number of elements, and Bob on a board with an odd number. Since Alice can never be forced to lose, she will continue playing until the board is small. Eventually, Bob will face a board with 1 element. He must remove it, making the XOR sum 0, and he will lose. Thus, if N is even, Alice wins.

• If nums.length is odd: Alice starts with an odd number of elements. She might be forced to lose on her first move (e.g., [7,7,7]). Even if she can make a non-losing move, she will leave a board with an even number of elements for Bob. Now, Bob is in the winning position described above. Bob has a strategy to never lose, so Alice will eventually lose. Thus, if N is odd, Alice loses.

Conclusion: Alice wins if initial_xor_sum == 0 OR nums.length % 2 == 0.

1class Solution {2    public boolean xorGame(int[] nums) {3        int xorSum = 0;4        for (int num : nums) {5            xorSum ^= num;6        }7 8        // Alice wins if the initial XOR sum is 0.9        if (xorSum == 0) {10            return true;11        }12 13        // If the XOR sum is not 0, the game depends on the number of elements.14        // If N is even, Alice can always make a move such that the XOR sum15        // of the remaining elements is not 0. Bob will be left in a state16        // with an odd number of elements. Eventually, Bob will be forced to lose.17        // If N is odd, Alice makes a move and leaves Bob with an even number of18        // elements, putting Bob in a winning position.19        return nums.length % 2 == 0;20    }21}