2315

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 '|'.

Companies: Google

Related Topics:
String

Solution 1.

// OJ: https://leetcode.com/problems/count-asterisks
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int countAsterisks(string s) {
        int ans = 0, N = s.size(), i = 0;
        for (; i < N; ) {
            while (i < N && s[i] != '|') {
                ans += s[i] == '*';
                ++i;
            }
            ++i;
            while (i < N && s[i] != '|') ++i;
            ++i;
        }
        return ans;
    }
};