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 <= 1000sconsists of lowercase English letters, vertical bars'|', and asterisks'*'.scontains an even number of vertical bars'|'.
Solution (Java)
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;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).