Added task 2315.

Author: javadev

README.md

@@ -1485,6 +1485,7 @@ implementation 'com.github.javadev:leetcode-in-java:1.11'
 
 | #    |      Title     | Difficulty  | Tag         | Time, ms | Time, %
 |------|----------------|-------------|-------------|----------|---------
+| 2315 |[Count Asterisks](src/main/java/g2301_2400/s2315_count_asterisks/Solution.java)| Easy || 1 | 100.00
 | 2312 |[Selling Pieces of Wood](src/main/java/g2301_2400/s2312_selling_pieces_of_wood/Solution.java)| Hard | Backtracking | 78 | 63.64
 | 2311 |[Longest Binary Subsequence Less Than or Equal to K](src/main/java/g2301_2400/s2311_longest_binary_subsequence_less_than_or_equal_to_k/Solution.java)| Medium | String, Dynamic_Programming, Greedy, Memoization | 1 | 100.00
 | 2310 |[Sum of Numbers With Units Digit K](src/main/java/g2301_2400/s2310_sum_of_numbers_with_units_digit_k/Solution.java)| Medium | Math | 1 | 66.67

src/main/java/g2301_2400/s2315_count_asterisks/Solution.java

@@ -0,0 +1,24 @@
+package g2301_2400.s2315_count_asterisks;
+
+// #Easy #2022_06_26_Time_1_ms_(100.00%)_Space_42.2_MB_(57.14%)
+
+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;
+    }
+}

src/main/java/g2301_2400/s2315_count_asterisks/readme.md

@@ -0,0 +1,43 @@
+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<sup>st</sup> and 2<sup>nd</sup> `'|'` make a pair, the 3<sup>rd</sup> and 4<sup>th</sup> `'|'` 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 `'|'`.

src/test/java/g2301_2400/s2315_count_asterisks/SolutionTest.java

@@ -0,0 +1,23 @@
+package g2301_2400.s2315_count_asterisks;
+
+import static org.hamcrest.CoreMatchers.equalTo;
+import static org.hamcrest.MatcherAssert.assertThat;
+
+import org.junit.jupiter.api.Test;
+
+class SolutionTest {
+    @Test
+    void countAsterisks() {
+        assertThat(new Solution().countAsterisks("l|*e*et|c**o|*de|"), equalTo(2));
+    }
+
+    @Test
+    void countAsterisks2() {
+        assertThat(new Solution().countAsterisks("iamprogrammer"), equalTo(0));
+    }
+
+    @Test
+    void countAsterisks3() {
+        assertThat(new Solution().countAsterisks("yo|uar|e**|b|e***au|tifu|l"), equalTo(5));
+    }
+}