Leetcode #2838: Maximum Coins Heroes Can Collect
In this guide, we solve Leetcode #2838 Maximum Coins Heroes Can Collect 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
There is a battle and n heroes are trying to defeat m monsters. You are given two 1-indexed arrays of positive integers heroes and monsters of length n and m, respectively.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Array, Two Pointers, Binary Search, Prefix Sum, 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.
Example
Input: heroes = [1,4,2], monsters = [1,1,5,2,3], coins = [2,3,4,5,6]
Output: [5,16,10]
Explanation: For each hero, we list the index of all the monsters he can defeat:
1st hero: [1,2] since the power of this hero is 1 and monsters[1], monsters[2] <= 1. So this hero collects coins[1] + coins[2] = 5 coins.
2nd hero: [1,2,4,5] since the power of this hero is 4 and monsters[1], monsters[2], monsters[4], monsters[5] <= 4. So this hero collects coins[1] + coins[2] + coins[4] + coins[5] = 16 coins.
3rd hero: [1,2,4] since the power of this hero is 2 and monsters[1], monsters[2], monsters[4] <= 2. So this hero collects coins[1] + coins[2] + coins[4] = 10 coins.
So the answer would be [5,16,10].
Python Solution
class Solution:
def maximumCoins(
self, heroes: List[int], monsters: List[int], coins: List[int]
) -> List[int]:
m = len(monsters)
idx = sorted(range(m), key=lambda i: monsters[i])
s = list(accumulate((coins[i] for i in idx), initial=0))
ans = []
for h in heroes:
i = bisect_right(idx, h, key=lambda i: monsters[i])
ans.append(s[i])
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.