Leetcode #2055: Plates Between Candles
In this guide, we solve Leetcode #2055 Plates Between Candles 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 long table with a line of plates and candles arranged on top of it. You are given a 0-indexed string s consisting of characters '' and '|' only, where a '' represents a plate and a '|' represents a candle.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, String, Binary Search, Prefix Sum
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: s = "**|**|***|", queries = [[2,5],[5,9]]
Output: [2,3]
Explanation:
- queries[0] has two plates between candles.
- queries[1] has three plates between candles.
Python Solution
class Solution:
def platesBetweenCandles(self, s: str, queries: List[List[int]]) -> List[int]:
n = len(s)
presum = [0] * (n + 1)
for i, c in enumerate(s):
presum[i + 1] = presum[i] + (c == '*')
left, right = [0] * n, [0] * n
l = r = -1
for i, c in enumerate(s):
if c == '|':
l = i
left[i] = l
for i in range(n - 1, -1, -1):
if s[i] == '|':
r = i
right[i] = r
ans = [0] * len(queries)
for k, (l, r) in enumerate(queries):
i, j = right[l], left[r]
if i >= 0 and j >= 0 and i < j:
ans[k] = presum[j] - presum[i + 1]
return ans
Complexity
The time complexity is O(log n) or O(n log n). The space complexity is O(1).
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.