How do you solve Plates Between Candles on LeetCode?

Plates Between Candles (LeetCode #2055) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + q log m) time complexity and O(n + m) space complexity. Key insight: Using prefix sums allows for quick calculations of counts within ranges.

What is the brute force solution for Plates Between Candles?

The brute force approach for Plates Between Candles is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks each query individually by extracting the substring and counting the plates between candles. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n + q log m).

What algorithm or data structure is used to solve Plates Between Candles?

Plates Between Candles uses the following concepts: Array, String, Binary Search, Prefix Sum. The recommended patterns to study are: Prefix Sum, Binary Search.

What are the interview tips for Plates Between Candles?

Always explain your thought process clearly; it shows understanding. Consider edge cases and test your solution with them. Practice writing clean and efficient code; it reflects your coding skills.

What are common mistakes when solving Plates Between Candles?

Not considering edge cases where there are no candles in the substring. Confusing the indices of plates and candles when counting.

#2055

Plates Between Candles

Medium

ArrayStringBinary SearchPrefix SumPrefix SumBinary Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + q log m)
Space	O(1)	O(n + m)

💡

Intuition

Time O(n + q log m)Space O(n + m)

By using prefix sums and pre-computed candle positions, we can efficiently answer each query in constant time after an initial linear pass through the string.

⚙️

Algorithm

4 steps

1Step 1: Create an array to store the prefix sum of plates up to each index.
2Step 2: Create an array to store the positions of all candles in the string.
3Step 3: For each query, use binary search to find the nearest candles to the left and right of the given indices.
4Step 4: Calculate the number of plates between these two candle indices using the prefix sum array.

solution.py21 lines

1def platesBetweenCandles(s, queries):
2    n = len(s)
3    prefix = [0] * n
4    candle_positions = []
5    count = 0
6    for i in range(n):
7        if s[i] == '*':
8            count += 1
9        prefix[i] = count
10        if s[i] == '|':
11            candle_positions.append(i)
12    results = []
13    for left, right in queries:
14        left_candle = bisect.bisect_right(candle_positions, left) - 1
15        right_candle = bisect.bisect_left(candle_positions, right)
16        if left_candle >= 0 and right_candle < len(candle_positions) and candle_positions[left_candle] < candle_positions[right_candle]:
17            plates_count = prefix[candle_positions[right_candle] - 1] - prefix[candle_positions[left_candle]]
18            results.append(plates_count)
19        else:
20            results.append(0)
21    return results

ℹ

Complexity note: The initial pass through the string is O(n) to create the prefix sums and candle positions, while each query can be answered in O(log m) due to binary search, where m is the number of candles.

1Using prefix sums allows for quick calculations of counts within ranges.
2Binary search helps efficiently locate the nearest candles for any given query.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.