LeetCode 2055: Plates Between Candles Solution

Master LeetCode problem 2055 (Plates Between Candles), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2055. Plates Between Candles

Medium

Array String Binary Search Prefix Sum

Problem Explanation

Explanation

To solve this problem, we can iterate through each query and count the number of plates that are between candles within the specified substring. We can do this by scanning the substring and checking if there is at least one candle to the left and at least one candle to the right of each plate.

We can maintain two arrays to store the positions of candles to the left and right of each plate in the substring.
For each query, we iterate through the substring and count the plates that have at least one candle to their left and right.
We return the count of plates meeting the criteria for each query.

Time Complexity:

The time complexity for this approach is O(N) per query, where N is the length of the substring.
Overall time complexity for all queries is O(Q*N), where Q is the number of queries and N is the maximum length of the substring.

Space Complexity:

The space complexity for this approach is O(N) to store the positions of candles to the left and right of each plate.

Solution Code

class Solution {
    public int[] countPlatesBetweenCandles(String s, int[][] queries) {
        int n = s.length();
        int[] leftCandle = new int[n];
        int[] rightCandle = new int[n];
        
        int left = -1;
        for (int i = 0; i < n; i++) {
            leftCandle[i] = left;
            if (s.charAt(i) == '|') {
                left = i;
            }
        }
        
        int right = -1;
        for (int i = n - 1; i >= 0; i--) {
            rightCandle[i] = right;
            if (s.charAt(i) == '|') {
                right = i;
            }
        }
        
        int[] result = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int leftIdx = queries[i][0];
            int rightIdx = queries[i][1];
            int count = 0;
            
            for (int j = leftIdx; j <= rightIdx; j++) {
                if (s.charAt(j) == '*') {
                    if (leftCandle[j] != -1 && rightCandle[j] != -1) {
                        count++;
                    }
                }
            }
            
            result[i] = count;
        }
        
        return result;
    }
}

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