LeetCode #2055 — MEDIUM

Plates Between Candles

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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] = [left_i, right_i] denotes the substring s[left_i...right_i] (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 i^th 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 <= left_i <= right_i < s.length

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

"**|**|***|"
[[2,5],[5,9]]

Example 2

"***|**|*****|**||**|*"
[[1,17],[4,5],[14,17],[5,11],[15,16]]

Core Insight

What unlocks the optimal approach

Can you find the indices of the most left and right candles for a given substring, perhaps by using binary search (or better) over an array of indices of all the bars?
Once the indices of the most left and right bars are determined, how can you efficiently count the number of plates within the range? Prefix sums?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2055: Plates Between Candles
class Solution {
    public int[] platesBetweenCandles(String s, int[][] queries) {
        int n = s.length();
        int[] presum = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            presum[i + 1] = presum[i] + (s.charAt(i) == '*' ? 1 : 0);
        }
        int[] left = new int[n];
        int[] right = new int[n];
        for (int i = 0, l = -1; i < n; ++i) {
            if (s.charAt(i) == '|') {
                l = i;
            }
            left[i] = l;
        }
        for (int i = n - 1, r = -1; i >= 0; --i) {
            if (s.charAt(i) == '|') {
                r = i;
            }
            right[i] = r;
        }
        int[] ans = new int[queries.length];
        for (int k = 0; k < queries.length; ++k) {
            int i = right[queries[k][0]];
            int j = left[queries[k][1]];
            if (i >= 0 && j >= 0 && i < j) {
                ans[k] = presum[j] - presum[i + 1];
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2055: Plates Between Candles
func platesBetweenCandles(s string, queries [][]int) []int {
	n := len(s)
	presum := make([]int, n+1)
	for i := range s {
		if s[i] == '*' {
			presum[i+1] = 1
		}
		presum[i+1] += presum[i]
	}
	left, right := make([]int, n), make([]int, n)
	for i, l := 0, -1; i < n; i++ {
		if s[i] == '|' {
			l = i
		}
		left[i] = l
	}
	for i, r := n-1, -1; i >= 0; i-- {
		if s[i] == '|' {
			r = i
		}
		right[i] = r
	}
	ans := make([]int, len(queries))
	for k, q := range queries {
		i, j := right[q[0]], left[q[1]]
		if i >= 0 && j >= 0 && i < j {
			ans[k] = presum[j] - presum[i+1]
		}
	}
	return ans
}

# Accepted solution for LeetCode #2055: Plates Between Candles
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

// Accepted solution for LeetCode #2055: Plates Between Candles
/**
 * [2055] Plates Between Candles
 *
 * 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 "*||<u>**</u>|". 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 i^th query.
 *  
 * Example 1:
 * <img alt="ex-1" src="https://assets.leetcode.com/uploads/2021/10/04/ex-1.png" style="width: 400px; height: 134px;" />
 * 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:
 * <img alt="ex-2" src="https://assets.leetcode.com/uploads/2021/10/04/ex-2.png" style="width: 600px; height: 193px;" />
 * 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 <= 10^5
 * 	s consists of '*' and '|' characters.
 * 	1 <= queries.length <= 10^5
 * 	queries[i].length == 2
 * 	0 <= lefti <= righti < s.length
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/plates-between-candles/
// discuss: https://leetcode.com/problems/plates-between-candles/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn plates_between_candles(s: String, queries: Vec<Vec<i32>>) -> Vec<i32> {
        vec![]
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2055_example_1() {
        let s = "**|**|***|".to_string();
        let queries = vec![vec![2, 5], vec![5, 9]];

        let result = vec![2, 3];

        assert_eq!(Solution::plates_between_candles(s, queries), result);
    }

    #[test]
    #[ignore]
    fn test_2055_example_2() {
        let s = "***|**|*****|**||**|*".to_string();
        let queries = vec![
            vec![1, 17],
            vec![4, 5],
            vec![14, 17],
            vec![5, 11],
            vec![15, 16],
        ];

        let result = vec![9, 0, 0, 0, 0];

        assert_eq!(Solution::plates_between_candles(s, queries), result);
    }
}

// Accepted solution for LeetCode #2055: Plates Between Candles
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2055: Plates Between Candles
// class Solution {
//     public int[] platesBetweenCandles(String s, int[][] queries) {
//         int n = s.length();
//         int[] presum = new int[n + 1];
//         for (int i = 0; i < n; ++i) {
//             presum[i + 1] = presum[i] + (s.charAt(i) == '*' ? 1 : 0);
//         }
//         int[] left = new int[n];
//         int[] right = new int[n];
//         for (int i = 0, l = -1; i < n; ++i) {
//             if (s.charAt(i) == '|') {
//                 l = i;
//             }
//             left[i] = l;
//         }
//         for (int i = n - 1, r = -1; i >= 0; --i) {
//             if (s.charAt(i) == '|') {
//                 r = i;
//             }
//             right[i] = r;
//         }
//         int[] ans = new int[queries.length];
//         for (int k = 0; k < queries.length; ++k) {
//             int i = right[queries[k][0]];
//             int j = left[queries[k][1]];
//             if (i >= 0 && j >= 0 && i < j) {
//                 ans[k] = presum[j] - presum[i + 1];
//             }
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.