LeetCode-in-Java

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 '|'.

Solution

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;
    }
}

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