Leetcode #948: Bag of Tokens
In this guide, we solve Leetcode #948 Bag of Tokens 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
You start with an initial power of power, an initial score of 0, and a bag of tokens given as an integer array tokens, where each tokens[i] denotes the value of tokeni. Your goal is to maximize the total score by strategically playing these tokens.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Array, Two Pointers, Sorting
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.
Python Solution
class Solution:
def bagOfTokensScore(self, tokens: List[int], power: int) -> int:
tokens.sort()
ans = score = 0
i, j = 0, len(tokens) - 1
while i <= j:
if power >= tokens[i]:
power -= tokens[i]
score, i = score + 1, i + 1
ans = max(ans, score)
elif score:
power += tokens[j]
score, j = score - 1, j - 1
else:
break
return ans
Complexity
The time complexity is , and the space complexity is . 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.