Leetcode #2868: The Wording Game
In this guide, we solve Leetcode #2868 The Wording Game in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Alice and Bob each have a lexicographically sorted array of strings named a and b respectively. They are playing a wording game with the following rules: On each turn, the current player should play a word from their list such that the new word is closely greater than the last played word; then it's the other player's turn.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Greedy, Array, Math, Two Pointers, String, Game Theory
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
Input: a = ["avokado","dabar"], b = ["brazil"]
Output: false
Explanation: Alice must start the game by playing the word "avokado" since it's her smallest word, then Bob plays his only word, "brazil", which he can play because its first letter, 'b', is the letter after Alice's word's first letter, 'a'.
Alice can't play a word since the first letter of the only word left is not equal to 'b' or the letter after 'b', 'c'.
So, Alice loses, and the game ends.
Python Solution
class Solution:
def canAliceWin(self, a: List[str], b: List[str]) -> bool:
i, j, k = 1, 0, 1
w = a[0]
while 1:
if k:
if j == len(b):
return True
if (b[j][0] == w[0] and b[j] > w) or ord(b[j][0]) - ord(w[0]) == 1:
w = b[j]
k ^= 1
j += 1
else:
if i == len(a):
return False
if (a[i][0] == w[0] and a[i] > w) or ord(a[i][0]) - ord(w[0]) == 1:
w = a[i]
k ^= 1
i += 1
Complexity
The time complexity is , where and are the lengths of arrays and , respectively. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.