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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2301_2400.s2315_count_asterisks;

// #Easy #String #2022_06_26_Time_1_ms_(100.00%)_Space_42.2_MB_(57.14%)

/**
 * 2315 - Count Asterisks\.
 *
 * Easy
 *
 * You are given a string `s`, where every **two** consecutive vertical bars `'|'` are grouped into a **pair**. In other words, the 1^st and 2^nd `'|'` make a pair, the 3^rd and 4^th `'|'` make a pair, and so forth.
 *
 * Return _the number of_ `'*'` _in_ `s`_, **excluding** the_ `'*'` _between each pair of_ `'|'`.
 *
 * **Note** that each `'|'` will belong to **exactly** one pair.
 *
 * **Example 1:**
 *
 * **Input:** s = "l|\*e\*et|c\*\*o|\*de|"
 *
 * **Output:** 2
 *
 * **Explanation:** The considered characters are underlined: "l|\*e\*et|c\*\*o|\*de|".
 *
 * The characters between the first and second '|' are excluded from the answer.
 *
 * Also, the characters between the third and fourth '|' are excluded from the answer. There are 2 asterisks considered. Therefore, we return 2.
 *
 * **Example 2:**
 *
 * **Input:** s = "iamprogrammer"
 *
 * **Output:** 0
 *
 * **Explanation:** In this example, there are no asterisks in s. Therefore, we return 0. 
 *
 * **Example 3:**
 *
 * **Input:** s = "yo|uar|e\*\*|b|e\*\*\*au|tifu|l"
 *
 * **Output:** 5
 *
 * **Explanation:** The considered characters are underlined: "yo|uar|e\*\*|b|e\*\*\*au|tifu|l".
 *
 * There are 5 asterisks considered. Therefore, we return 5.
 *
 * **Constraints:**
 *
 * *   `1 <= s.length <= 1000`
 * *   `s` consists of lowercase English letters, vertical bars `'|'`, and asterisks `'*'`.
 * *   `s` contains an **even** number of vertical bars `'|'`.
**/
public class Solution {
    public int countAsterisks(String s) {
        int c = 0;
        int n = s.length();
        int i = 0;
        while (i < n) {
            if (s.charAt(i) == '|') {
                i++;
                while (s.charAt(i) != '|') {
                    i++;
                }
            }
            if (s.charAt(i) == '*') {
                c++;
            }
            i++;
        }
        return c;
    }
}