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Formatted question description: https://leetcode.ca/all/2315.html

2315. Count Asterisks

  • Difficulty: Easy.
  • Related Topics: String.
  • Similar Questions: .

Problem

You are given a string s, where every two consecutive vertical bars '|' are grouped into a pair. In other words, the 1st and 2nd '|' make a pair, the 3rd and 4th '|' 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 (Java, C++, Python)

  • 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;
    }
}

############

class Solution {
    public int countAsterisks(String s) {
        int ans = 0;
        for (int i = 0, ok = 1; i < s.length(); ++i) {
            char c = s.charAt(i);
            if (c == '*') {
                ans += ok;
            } else if (c == '|') {
                ok ^= 1;
            }
        }
        return ans;
    }
}

  • class Solution:
    def countAsterisks(self, s: str) -> int:
        ans, ok = 0, 1
        for c in s:
            if c == "*":
                ans += ok
            elif c == "|":
                ok ^= 1
        return ans

############

# 2315. Count Asterisks
# https://leetcode.com/problems/count-asterisks/

class Solution:
    def countAsterisks(self, s: str) -> int:
        A = s.split('|')
        n = len(A)
        res = 0
        
        for i in range(0, n, 2):
            res += A[i].count("*")
        
        return res


  • class Solution {
public:
    int countAsterisks(string s) {
        int ans = 0, ok = 1;
        for (char& c : s) {
            if (c == '*') {
                ans += ok;
            } else if (c == '|') {
                ok ^= 1;
            }
        }
        return ans;
    }
};

  • func countAsterisks(s string) (ans int) {
	ok := 1
	for _, c := range s {
		if c == '*' {
			ans += ok
		} else if c == '|' {
			ok ^= 1
		}
	}
	return
}

  • function countAsterisks(s: string): number {
    let ans = 0;
    let ok = 1;
    for (const c of s) {
        if (c === '*') {
            ans += ok;
        } else if (c === '|') {
            ok ^= 1;
        }
    }
    return ans;
}


  • public class Solution {
    public int CountAsterisks(string s) {
        int ans = 0, ok = 1;
        foreach (char c in s) {
            if (c == '*') {
                ans += ok;
            } else if (c == '|') {
                ok ^= 1;
            }
        }
        return ans;
    }
}

  • impl Solution {
    pub fn count_asterisks(s: String) -> i32 {
        let mut ans = 0;
        let mut ok = 1;
        for &c in s.as_bytes() {
            if c == b'*' {
                ans += ok
            } else if c == b'|' {
                ok ^= 1
            }
        }
        ans
    }
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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