LeetCode 2315: Count Asterisks Solution

Problem Explanation

Explanation

To solve this problem, we iterate through the string s and count the number of asterisks that are not between pairs of vertical bars. We can achieve this by keeping track of whether we are inside a pair of vertical bars or not.

Initialize a variable count to 0 to store the count of asterisks.
Initialize a boolean variable insideBars to false to indicate whether we are inside a pair of vertical bars or not.
Iterate through each character in the string s.
- If the current character is a vertical bar '|':
  - Toggle the value of insideBars.
- If the current character is an asterisk '*' and we are not inside a pair of vertical bars:
  - Increment the count.
Return the final count.

Time Complexity

The time complexity of this algorithm is O(n), where n is the length of the input string s.

Space Complexity

The space complexity of this algorithm is O(1) as we are using a constant amount of extra space.

Solution Code

class Solution {
    public int countAsterisks(String s) {
        int count = 0;
        boolean insideBars = false;

        for (char c : s.toCharArray()) {
            if (c == '|') {
                insideBars = !insideBars;
            } else if (c == '*' && !insideBars) {
                count++;
            }
        }

        return count;
    }
}

Frequently Asked Questions

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2315. Count Asterisks