Leetcode #2921: Maximum Profitable Triplets With Increasing Prices II
In this guide, we solve Leetcode #2921 Maximum Profitable Triplets With Increasing Prices II 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
Given the 0-indexed arrays prices and profits of length n. There are n items in an store where the ith item has a price of prices[i] and a profit of profits[i].
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Binary Indexed Tree, Segment Tree, Array
Intuition
The constraints allow a direct scan that keeps only the essential state.
By translating the requirements into a clean loop, the logic stays easy to reason about.
Approach
Iterate through the data once, updating the state needed to compute the answer.
Return the final state after the traversal is complete.
Steps:
- Parse the input.
- Iterate and update state.
- Return the computed answer.
Example
Input: prices = [10,2,3,4], profits = [100,2,7,10]
Output: 19
Explanation: We can't pick the item with index i=0 since there are no indices j and k such that the condition holds.
So the only triplet we can pick, are the items with indices 1, 2 and 3 and it's a valid pick since prices[1] < prices[2] < prices[3].
The answer would be sum of their profits which is 2 + 7 + 10 = 19.
Python Solution
class BinaryIndexedTree:
def __init__(self, n: int):
self.n = n
self.c = [0] * (n + 1)
def update(self, x: int, v: int):
while x <= self.n:
self.c[x] = max(self.c[x], v)
x += x & -x
def query(self, x: int) -> int:
mx = 0
while x:
mx = max(mx, self.c[x])
x -= x & -x
return mx
class Solution:
def maxProfit(self, prices: List[int], profits: List[int]) -> int:
n = len(prices)
left = [0] * n
right = [0] * n
m = max(prices)
tree1 = BinaryIndexedTree(m + 1)
tree2 = BinaryIndexedTree(m + 1)
for i, x in enumerate(prices):
left[i] = tree1.query(x - 1)
tree1.update(x, profits[i])
for i in range(n - 1, -1, -1):
x = m + 1 - prices[i]
right[i] = tree2.query(x - 1)
tree2.update(x, profits[i])
return max(
(l + x + r for l, x, r in zip(left, profits, right) if l and r), default=-1
)
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.