2315. Count Asterisks

Description

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 '|'.

Solutions

Solution 1: Simulation

We define an integer variable \(\textit{ok}\) to indicate whether we can count when encountering *. Initially, \(\textit{ok}=1\), meaning we can count.

Traverse the string \(s\). If we encounter *, we decide whether to count based on the value of \(\textit{ok}\). If we encounter |, we toggle the value of \(\textit{ok}\).

Finally, return the count result.

The time complexity is \(O(n)\), where \(n\) is the length of the string \(s\). The space complexity is \(O(1)\).

Python3JavaC++GoTypeScriptRustC#C

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

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 {
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;
}

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
    }
}

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;
    }
}

int countAsterisks(char* s) {
    int ans = 0;
    int ok = 1;
    for (int i = 0; s[i]; i++) {
        if (s[i] == '*') {
            ans += ok;
        } else if (s[i] == '|') {
            ok ^= 1;
        }
    }
    return ans;
}

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