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2678. Number of Senior Citizens

Description

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.

Solutions

Solution 1: Traversal and Counting

We can traverse each string $x$ in details and convert the $12$th and $13$th characters (indexed at $11$ and $12$) of $x$ to integers, and check if they are greater than $60$. If so, we add one to the answer.

After the traversal, we return the answer.

The time complexity is $O(n)$, where $n$ is the length of details. The space complexity is $O(1)`.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public int countSeniors(String[] details) {
          int ans = 0;
          for (var x : details) {
              int age = Integer.parseInt(x.substring(11, 13));
              if (age > 60) {
                  ++ans;
              }
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int countSeniors(vector<string>& details) {
          int ans = 0;
          for (auto& x : details) {
              int age = stoi(x.substr(11, 2));
              ans += age > 60;
          }
          return ans;
      }
  };
  

• class Solution:
      def countSeniors(self, details: List[str]) -> int:
          return sum(int(x[11:13]) > 60 for x in details)
  
  

• func countSeniors(details []string) (ans int) {
  	for _, x := range details {
  		age, _ := strconv.Atoi(x[11:13])
  		if age > 60 {
  			ans++
  		}
  	}
  	return
  }
  

• function countSeniors(details: string[]): number {
      let ans = 0;
      for (const x of details) {
          const age = parseInt(x.slice(11, 13));
          if (age > 60) {
              ++ans;
          }
      }
      return ans;
  }
  
  

• impl Solution {
      pub fn count_seniors(details: Vec<String>) -> i32 {
          let mut ans = 0;
  
          for s in details.iter() {
              if let Ok(age) = s[11..13].parse::<i32>() {
                  if age > 60 {
                      ans += 1;
                  }
              }
          }
  
          ans
      }
  }
  
  

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