Leetcode #1475: Final Prices With a Special Discount in a Shop
In this guide, we solve Leetcode #1475 Final Prices With a Special Discount in a Shop 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 are given an integer array prices where prices[i] is the price of the ith item in a shop. There is a special discount for items in the shop.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Stack, Array, Monotonic Stack
Intuition
We need the next greater or smaller element efficiently, which is exactly what a monotonic stack offers.
Each element is pushed and popped at most once, yielding a linear-time scan.
Approach
Maintain a stack that is either increasing or decreasing, depending on the query.
When the invariant is broken, pop and resolve answers for those indices.
Steps:
- Scan elements once.
- Pop while the monotonic condition is violated.
- Use stack indices to update answers.
Example
Input: prices = [8,4,6,2,3]
Output: [4,2,4,2,3]
Explanation:
For item 0 with price[0]=8 you will receive a discount equivalent to prices[1]=4, therefore, the final price you will pay is 8 - 4 = 4.
For item 1 with price[1]=4 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 4 - 2 = 2.
For item 2 with price[2]=6 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 6 - 2 = 4.
For items 3 and 4 you will not receive any discount at all.
Python Solution
class Solution:
def finalPrices(self, prices: List[int]) -> List[int]:
stk = []
for i in reversed(range(len(prices))):
x = prices[i]
while stk and x < stk[-1]:
stk.pop()
if stk:
prices[i] -= stk[-1]
stk.append(x)
return prices
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.