# Solving Leetcode Interviews in Seconds with AI: Plates Between Candles


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2055" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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. You are also given a 0-indexed 2D integer array queries where queries[i] = [lefti, righti] denotes the substring s[lefti...righti] (inclusive). For each query, you need to find the number of plates between candles that are in the substring. A plate is considered between candles if there is at least one candle to its left and at least one candle to its right in the substring.  For example, s = "||**||**|*", and a query [3, 8] denotes the substring "*||**|". The number of plates between candles in this substring is 2, as each of the two plates has at least one candle in the substring to its left and right.  Return an integer array answer where answer[i] is the answer to the ith query.   Example 1:   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.  Example 2:   Input: s = "***|**|*****|**||**|*", queries = [[1,17],[4,5],[14,17],[5,11],[15,16]] Output: [9,0,0,0,0] Explanation: - 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  

	# Explanation
	Here's a breakdown of the solution approach, followed by the code:

*   **Prefix Sum for Plates:** Calculate a prefix sum array `plates` where `plates[i]` stores the number of plates up to index `i` in the string `s`. This allows us to efficiently count the plates within a substring.
*   **Nearest Candles:** Create two arrays, `left` and `right`. `left[i]` stores the index of the nearest candle to the left of index `i` (inclusive), and `right[i]` stores the index of the nearest candle to the right of index `i` (inclusive).
*   **Query Processing:** For each query `[lefti, righti]`, find the nearest candles within the range using `left` and `right` arrays. Calculate the number of plates between these candles using the `plates` prefix sum.

*   **Runtime Complexity:** O(n + q), where n is the length of the string `s` and q is the number of queries.
*   **Storage Complexity:** O(n) due to the `plates`, `left`, and `right` arrays.

	
	# Code
	```python
	def plates_between_candles(s: str, queries: list[list[int]]) -> list[int]:
    n = len(s)
    plates = [0] * n
    left = [0] * n
    right = [0] * n
    
    # Calculate prefix sum of plates
    plates[0] = 1 if s[0] == '*' else 0
    for i in range(1, n):
        plates[i] = plates[i - 1] + (1 if s[i] == '*' else 0)
    
    # Find nearest candle to the left
    last_candle = -1
    for i in range(n):
        if s[i] == '|':
            last_candle = i
        left[i] = last_candle
    
    # Find nearest candle to the right
    last_candle = -1
    for i in range(n - 1, -1, -1):
        if s[i] == '|':
            last_candle = i
        right[i] = last_candle
    
    # Process queries
    result = []
    for l, r in queries:
        left_candle = right[l]
        right_candle = left[r]
        
        if left_candle == -1 or right_candle == -1 or left_candle >= right_candle:
            result.append(0)
        else:
            result.append(plates[right_candle] - plates[left_candle])
    
    return result
	```