Number of Senior Citizens - Practice Coding Problems

Number of Senior Citizens - Problem

Array String Easy

You're working as a data analyst for an airline company and need to identify senior citizens from passenger records for special assistance services.

Given a 0-indexed array of strings called details, where each string contains compressed passenger information in a fixed 15-character format:

Characters 0-9: Phone number (10 digits)
Character 10: Gender ('M' or 'F')
Characters 11-12: Age (2 digits, e.g., '65', '42')
Characters 13-14: Seat number (2 characters)

Goal: Return the count of passengers who are strictly older than 60 years (age > 60).

Example: If a passenger string is "7868190130M7522", the age is 75 (characters 11-12), so this passenger counts as a senior citizen.

Input & Output

example_1.py — Basic Case

$ Input: ["7868190130M7522", "5303914400F9211", "9273338640F4823"]

› Output: 2

💡 Note: First passenger: age 75 > 60 ✓, Second passenger: age 92 > 60 ✓, Third passenger: age 48 ≤ 60 ✗. Total: 2 seniors

example_2.py — Edge Case

$ Input: ["1313579440F2036", "2921522980M5644"]

› Output: 0

💡 Note: First passenger: age 20 ≤ 60, Second passenger: age 56 ≤ 60. No passengers are over 60, so count is 0

example_3.py — Boundary Case

$ Input: ["9751302862F0693", "3888560693F7262", "5485983835F0649"]

› Output: 1

💡 Note: Ages are 06, 72, and 06 respectively. Only the second passenger (age 72) is over 60. Count is 1

Constraints

1 ≤ details.length ≤ 500
details[i].length == 15
details[i] consists of digits and an uppercase English letter
All phone numbers and seat numbers are distinct
Age is always a valid 2-digit number (00-99)

Visualization

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Understanding the Visualization

Setup Scanner

Initialize counter and prepare to scan boarding passes

Scan Pass

Read the 15-character boarding pass data

Locate Age

Extract digits from positions 11-12 (age section)

Check Senior Status

Convert age to number and check if > 60

Update Count

Increment senior citizen counter if applicable

Key Takeaway

🎯 Key Insight: String manipulation with fixed positions allows for direct, efficient data extraction - no complex algorithms needed, just precise indexing!

Asked in

a Amazon 15 ⊞ Microsoft 8 G Google 5 🍎 Apple 3

The optimal solution uses direct string processing to extract the age from positions 11-12 of each passenger string. This approach achieves O(n) time complexity with O(1) space complexity, making it both time and space efficient. The key insight is that we need to examine every passenger anyway, so a single pass with substring extraction is the most straightforward and efficient method.

Common Approaches

Approach	Time	Space	Notes
✓ Direct String Processing	O(n)	O(1)	Extract age from each passenger string and count seniors

Direct String Processing — Algorithm Steps

Initialize a counter variable to 0
Loop through each passenger detail string
Extract age substring from positions 11-12
Convert age string to integer
If age > 60, increment counter
Return the final count

Visualization

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Step-by-Step Walkthrough

Initialize Counter

Start with count = 0

Scan Each String

Process each 15-character passenger string

Extract Age

Get characters at positions 11-12

Check Age

Convert to integer and compare with 60

Update Counter

Increment count if age > 60

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int countSeniors(char details[][16], int size) {
    int count = 0;
    for (int i = 0; i < size; i++) {
        // Extract age from positions 11-12
        char ageStr[3];
        ageStr[0] = details[i][11];
        ageStr[1] = details[i][12];
        ageStr[2] = '\0';
        
        int age = atoi(ageStr);
        if (age > 60) {
            count++;
        }
    }
    return count;
}

int main() {
    char details[100][16];
    int size = 0;
    
    // Simple input reading for demo
    char input[1000];
    fgets(input, sizeof(input), stdin);
    
    char* token = strtok(input, "\"");
    while (token != NULL && size < 100) {
        if (strlen(token) == 15) {
            strcpy(details[size], token);
            size++;
        }
        token = strtok(NULL, "\"");
    }
    
    printf("%d\n", countSeniors(details, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We must check every passenger detail exactly once, where n is the number of passengers

✓ Linear Growth

Space Complexity

O(1)

Only using a constant amount of extra space for the counter and temporary variables

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Direct String Processing — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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