2678. Number of Senior Citizens

You are given a 0-indexed array of strings details. Each element of details provides information about a given passenger compressed into a string of length 15. The system is such that:

• The first ten characters consist of the phone number of passengers.
• The next character denotes the gender of the person.
• The following two characters are used to indicate the age of the person.
• The last two characters determine the seat allotted to that person.

Return the number of passengers who are strictly more than 60 years old.

Example 1:

Input: details = ["7868190130M7522","5303914400F9211","9273338290F4010"]
Output: 2
Explanation: The passengers at indices 0, 1, and 2 have ages 75, 92, and 40. Thus, there are 2 people who are over 60 years old.

Example 2:

Input: details = ["1313579440F2036","2921522980M5644"]
Output: 0
Explanation: None of the passengers are older than 60.

Constraints:

• 1 <= details.length <= 100
• details[i].length == 15
• details[i] consists of digits from '0' to '9'.
• details[i][10] is either 'M' or 'F' or 'O'.
• The phone numbers and seat numbers of the passengers are distinct.

Approach 01:

#include <vector>
#include <string>

using namespace std;

class Solution {
public:
    int countSeniors(vector<string>& details) {
        int seniorCount = 0;
        for (const string& detail : details) {
            // Extract the age from the last 4th and 3rd characters and convert to integer
            int age = stoi(detail.substr(detail.size() - 4, 2));
            if (age > 60) {
                seniorCount += 1;
            }
        }
        return seniorCount;
    }
};

class Solution:
    def countSeniors(self, details: List[str]) -> int:
        seniorCount = 0
        for detail in details:
            age = int(detail[-4:-2])
            if age > 60:
                seniorCount += 1
        return seniorCount

Time Complexity

• Iterating through the list of details:

  The algorithm iterates through each string in the details vector. If there are \( n \) strings in the vector, this step takes \( O(n) \) time.

• Extracting the age from each string:

  For each string, extracting the age involves the substr and stoi functions, which take constant time \( O(1) \). Since this operation is done for each of the \( n \) strings, the total time for this step is also \( O(n) \).

• Overall Time Complexity:

  The overall time complexity is \( O(n) \).

Space Complexity

• Auxiliary Space:

  The space used by the seniorCount variable and the age variable is constant, i.e., \( O(1) \).

• Overall Space Complexity:

  The overall space complexity is \( O(1) \) as the space used does not depend on the size of the input vector.