#2055

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Plates Between Candles

Determine the number of plates between candles in a given substring using binary search and prefix sum techniques.

Problem Statement

You are given a string s consisting of '*' representing plates and '|' representing candles. Your task is to answer multiple queries on substrings of s, each querying how many plates lie between two candles in a specific range.

A plate is considered between candles if there is at least one candle on either side of it. You need to return an integer array answer where each element corresponds to the number of plates between candles for the corresponding query in the input queries array.

Examples

Example 1

Input: s = "||*|", queries = [[2,5],[5,9]]

Output: [2,3]

queries[0] has two plates between candles.
queries[1] has three plates between candles.

Example 2

Input: s = "*||*||||*", queries = [[1,17],[4,5],[14,17],[5,11],[15,16]]

Output: [9,0,0,0,0]

queries[0] has nine plates between candles.
The other queries have zero plates between candles.

Constraints

3 <= s.length <= 105
s consists of '*' and '|' characters.
1 <= queries.length <= 105
queries[i].length == 2
0 <= lefti <= righti < s.length

Solution Approach

Prefix Sum Array for Plates

To quickly count the plates in any substring, use a prefix sum array to store cumulative plate counts up to each index. This allows you to calculate the number of plates between any two indices in constant time.

Binary Search for Candle Indices

For each query, locate the nearest left and right candles using binary search on the indices of candles in the string. This ensures you can quickly identify valid segments where plates are surrounded by candles.

Efficient Query Handling

By combining the prefix sum array and binary search to find candle positions, you can handle each query in logarithmic time, making the approach efficient enough to handle the problem's constraints.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity for handling each query is O(log N) due to binary search, and the preprocessing step of building the prefix sum array takes O(N). Thus, the overall time complexity is O(N + Q log N), where N is the length of the string and Q is the number of queries. Space complexity is O(N) due to the prefix sum and candle position arrays.

What Interviewers Usually Probe

Look for understanding of binary search over answer space and efficient query handling.
Test the candidate's ability to optimize the problem using prefix sums.
Check for the candidate's familiarity with handling large inputs efficiently.

Common Pitfalls or Variants

Common pitfalls

Incorrectly handling queries where no valid plate is between candles.
Failing to optimize the solution for large input sizes, leading to timeouts.
Misunderstanding how to handle cases where candles are adjacent or missing in the query range.

Follow-up variants

Handling cases where no candles are present in the substring.
Optimizing for multiple queries with overlapping ranges.
Modifying the problem to count plates without requiring both left and right candles.

FAQ

How can I optimize the 'Plates Between Candles' problem for multiple queries?

The key optimization is using binary search for candle positions and a prefix sum array to quickly calculate plate counts within a range.

What is the role of prefix sums in solving the 'Plates Between Candles' problem?

Prefix sums allow for quick plate counting between any two indices, reducing the time complexity of the solution.

What is the time complexity of solving the 'Plates Between Candles' problem?

The time complexity is O(N + Q log N), where N is the string length and Q is the number of queries.

Can I solve the 'Plates Between Candles' problem without using binary search?

While it's possible, binary search significantly improves the efficiency by quickly locating the nearest candles, making it crucial for larger inputs.

What are some common pitfalls in the 'Plates Between Candles' problem?

Common pitfalls include incorrectly handling edge cases where no candles are present in the query range or when candles are adjacent.

terminal

Solution

Solution 1

#### Python3

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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

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