Problem Description

You have a linked list with an even number of nodes. The nodes are indexed starting from 0, where:

• Nodes at even indices (0, 2, 4, ...) contain even integers
• Nodes at odd indices (1, 3, 5, ...) contain odd integers

The list is divided into pairs of consecutive nodes:

• Nodes at indices 0 and 1 form the first pair
• Nodes at indices 2 and 3 form the second pair
• And so on...

For each pair, you need to compare the values:

• If the odd-indexed node (second node in the pair) has a higher value, the "Odd" team gets a point
• If the even-indexed node (first node in the pair) has a higher value, the "Even" team gets a point

Your task is to determine which team wins by returning:

• "Odd" if the Odd team has more points
• "Even" if the Even team has more points
• "Tie" if both teams have equal points

For example, given the list [2,5,4,7,20,5]:

• Pair (2,5): Since 5 > 2, Odd team gets 1 point
• Pair (4,7): Since 7 > 4, Odd team gets 1 point
• Pair (20,5): Since 20 > 5, Even team gets 1 point
• Final score: Odd team has 2 points, Even team has 1 point, so return "Odd"

The solution traverses the linked list in steps of two nodes, comparing each pair and keeping track of the scores. After processing all pairs, it compares the final scores to determine the winner.

Intuition

The problem is essentially asking us to compare pairs of consecutive nodes and keep score. Since we're dealing with a linked list where every two consecutive nodes form a pair, the natural approach is to traverse the list two nodes at a time.

The key insight is that we don't need to track node indices explicitly. Since the list has an even length and we're told that even-indexed nodes contain even values while odd-indexed nodes contain odd values, we can simply process the list in chunks of two nodes. The first node we encounter in each chunk is always at an even index, and the second is at an odd index.

For scoring, we need two counters - one for the "Odd" team and one for the "Even" team. As we process each pair:

• If the first node's value is less than the second node's value (a < b), the odd-indexed node wins, so we increment the odd counter
• If the first node's value is greater than the second node's value (a > b), the even-indexed node wins, so we increment the even counter

The elegant part of the solution is using boolean expressions directly: odd += a < b and even += a > b. In Python, True evaluates to 1 and False evaluates to 0, so these expressions naturally increment the counters only when the respective condition is met.

After traversing all pairs, we simply compare the two scores to determine the winner. This straightforward simulation approach works because we're just following the rules of the game exactly as described - no complex transformations or algorithms needed.

Learn more about Linked List patterns.

Solution Approach

The solution uses a simple simulation approach by traversing the linked list and processing pairs of nodes sequentially.

Implementation Steps:

1. Initialize counters: Create two variables odd and even, both set to 0, to track the scores for each team.

2. Traverse the list in pairs: Use a while loop that continues as long as head is not null. In each iteration:

   • Extract the value of the current node: a = head.val
   • Extract the value of the next node: b = head.next.val
   • These two nodes form a pair where a is at an even index and b is at an odd index

3. Update scores based on comparison:

   • odd += a < b: If the odd-indexed node (b) is greater than the even-indexed node (a), increment the odd team's score
   • even += a > b: If the even-indexed node (a) is greater than the odd-indexed node (b), increment the even team's score
   • Note: Python treats True as 1 and False as 0, so these boolean expressions directly add 1 or 0 to the counters

4. Move to the next pair: Advance the pointer by two nodes using head = head.next.next to reach the start of the next pair.

5. Determine the winner: After processing all pairs, compare the final scores:

   • If odd > even, return "Odd"
   • If odd < even, return "Even"
   • If scores are equal, return "Tie"

Time Complexity: O(n) where n is the number of nodes in the linked list, as we visit each node exactly once.

Space Complexity: O(1) as we only use two integer variables to track scores, regardless of the input size.

The beauty of this solution lies in its simplicity - it directly simulates the game rules without any unnecessary complexity, making the code both efficient and easy to understand.

Example Walkthrough

Let's walk through a small example with the linked list: [6, 3, 10, 1]

Initial Setup:

• odd = 0 (score for odd team)
• even = 0 (score for even team)
• head points to the first node (value 6)

First Iteration:

• Current pair: nodes at indices 0 and 1
• a = head.val = 6 (even index, even value)
• b = head.next.val = 3 (odd index, odd value)
• Compare: Is 6 < 3? No, so odd += False (odd remains 0)
• Compare: Is 6 > 3? Yes, so even += True (even becomes 1)
• Move forward: head = head.next.next (now points to node with value 10)
• Scores: odd = 0, even = 1

Second Iteration:

• Current pair: nodes at indices 2 and 3
• a = head.val = 10 (even index, even value)
• b = head.next.val = 1 (odd index, odd value)
• Compare: Is 10 < 1? No, so odd += False (odd remains 0)
• Compare: Is 10 > 1? Yes, so even += True (even becomes 2)
• Move forward: head = head.next.next (now becomes null)
• Scores: odd = 0, even = 2

Determine Winner:

• Final scores: odd = 0, even = 2
• Since even > odd, return "Even"

The Even team wins with 2 points to 0, having won both pair comparisons (6 > 3 and 10 > 1).

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the linked list. The algorithm traverses the linked list once, visiting each node exactly once. In each iteration of the while loop, we process two nodes (the current node and its next node) and then move forward by two positions with head = head.next.next. Since we visit all n nodes in the list exactly once, the time complexity is linear.

Space Complexity: O(1). The algorithm uses only a constant amount of extra space regardless of the input size. We only maintain two integer variables (odd and even) to keep track of the scores, plus a few temporary variables (a and b) to store node values. These variables do not grow with the size of the input linked list, resulting in constant space complexity.

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

Common Pitfalls

1. Null Pointer Exception When Accessing head.next.val

The most critical pitfall in this solution is assuming that head.next always exists when processing pairs. If someone modifies the code or uses it in a different context, accessing head.next.val without checking if head.next is null will cause a runtime error.

Problem Example:

# This will crash if head.next is None
second_value = head.next.val  # AttributeError if head.next is None

Solution: Add a safety check before accessing the next node:

while head and head.next:  # Check both head and head.next
    first_value = head.val
    second_value = head.next.val
    # ... rest of the logic

2. Incorrect Handling of Odd Number of Nodes

While the problem states the list has an even number of nodes, defensive programming suggests handling edge cases. If the list accidentally has an odd number of nodes, the current implementation will crash.

Solution: Either validate the input or handle incomplete pairs:

def gameResult(self, head: Optional[ListNode]) -> str:
    odd_score = 0
    even_score = 0
  
    # Handle edge case of empty list
    if not head:
        return "Tie"
  
    while head and head.next:
        first_value = head.val
        second_value = head.next.val
      
        odd_score += first_value < second_value
        even_score += first_value > second_value
      
        head = head.next.next
  
    # Optional: Check if there's a leftover node (odd length list)
    if head:
        # Could handle this case or raise an exception
        pass
  
    if odd_score > even_score:
        return "Odd"
    if odd_score < even_score:
        return "Even"
    return "Tie"

3. Confusion About Index-Based Team Names

The problem states that nodes at even indices contain even integers and nodes at odd indices contain odd integers, but the team names ("Odd" and "Even") refer to the index positions, not the values themselves. This can lead to confusion when reading or modifying the code.

Solution: Use clearer variable names and add comments:

# More descriptive variable names
odd_index_team_score = 0   # Score for nodes at odd indices (1, 3, 5...)
even_index_team_score = 0  # Score for nodes at even indices (0, 2, 4...)

# Or use position-based naming
second_position_wins = 0  # When second node in pair wins
first_position_wins = 0   # When first node in pair wins

4. Integer Overflow in Score Counters

While unlikely in Python (which handles arbitrary precision integers), in other languages with fixed-size integers, very long lists could cause integer overflow in the score counters.

Solution for other languages: Use appropriate data types or implement overflow checking:

# In languages with fixed integers, consider using long/int64
# or checking for overflow conditions
if odd_score == sys.maxsize:  # Example overflow check
    # Handle overflow case
    pass