2055. Plates Between Candles
Description
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 is2
, 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
Solutions
Solution: Binary Search + Prefix Sum
- Time complexity: O(nlogn)
- Space complexity: O(n)
JavaScript
js
/**
* @param {string} s
* @param {number[][]} queries
* @return {number[]}
*/
const platesBetweenCandles = function (s, queries) {
const PLATE = '*';
const prefixCandles = [];
const findCandle = (left, right, target) => {
while (left < right) {
const middle = Math.floor((left + right) / 2);
prefixCandles[middle] >= target ? (right = middle) : (left = middle + 1);
}
return prefixCandles[left] === target ? left : -1;
};
for (const [index, element] of s.entries()) {
const previous = prefixCandles[index - 1] ?? 0;
prefixCandles[index] = element === PLATE ? previous : previous + 1;
}
return queries.map(([left, right]) => {
const leftCandle = s[left] === PLATE ? findCandle(left, right, prefixCandles[left] + 1) : left;
if (leftCandle === -1) return 0;
const rightCandle = s[right] === PLATE ? findCandle(left, right, prefixCandles[right]) : right;
if (rightCandle === -1) return 0;
const candles = prefixCandles[rightCandle] - prefixCandles[leftCandle] + 1;
return rightCandle - leftCandle + 1 - candles;
});
};