27. Convert Seconds to Time Format

📝 Problem

Write a program to read the total seconds and print the seconds in time format Example :hr:min:sec

Input Format:
Accept integer as a input

Output Format:
Display the output in the format "**:HOUR **:MIN :**SEC"

Constraints:
1 ≤ sec ≤ 10^15

Sample Input 1:

54300

Sample Output 1:

15:HOUR 05:MIN :00SEC

Sample Input 2:

76000

Sample Output 2:

21:HOUR 06:MIN :40SEC

🎯 Time Conversion: Mastering Division & Modulo Operations!

The Challenge: Breaking Down Time Units

Converting total seconds into hours, minutes, and seconds is a classic problem that teaches you how to decompose a value into multiple units. This is fundamental to understanding time, currency, and any hierarchical unit system!

💡 Why This Question is Special

This problem demonstrates the power of division (/) and modulo (%) operators working together. You'll learn how to extract different "layers" of information from a single number - a technique used everywhere from time calculations to currency conversions!

1️⃣ Understanding Time Units

The Time Hierarchy:

📌 Time Unit Relationships:

1 minute  = 60 seconds
1 hour    = 60 minutes = 3600 seconds
1 day     = 24 hours   = 86400 seconds

// So for 54300 seconds:
// How many full hours? 54300 ÷ 3600 = 15 hours
// Remaining seconds? 54300 % 3600 = 300 seconds
// How many full minutes? 300 ÷ 60 = 5 minutes
// Remaining seconds? 300 % 60 = 0 seconds

2️⃣ The Algorithm: Step-by-Step Breakdown

📌 Example with 54300 seconds:

long long totalSec = 54300;

// Step 1: Extract HOURS
long long hours = totalSec / 3600;
// 54300 / 3600 = 15 hours

// Step 2: Get remaining seconds after removing hours
totalSec = totalSec % 3600;
// 54300 % 3600 = 300 seconds left

// Step 3: Extract MINUTES from remaining seconds
long long minutes = totalSec / 60;
// 300 / 60 = 5 minutes

// Step 4: Get remaining SECONDS
long long seconds = totalSec % 60;
// 300 % 60 = 0 seconds

// Result: 15:HOUR 05:MIN :00SEC ✓

3️⃣ The Magic of Division (/) and Modulo (%)

Operation	Symbol	Purpose	Example
Division	`/`	Counts how many times divisor fits	`54300 / 3600 = 15`
Modulo	`%`	Gets the remainder	`54300 % 3600 = 300`

💡 The Perfect Partnership

Division (/) and Modulo (%) work together like a team:

/ tells you "how many complete units"
% tells you "what's left over"

Together, they let you decompose any value into hierarchical components!

4️⃣ Visual Breakdown of 76000 Seconds

📌 Let's Trace Through Example 2:

// Input: 76000 seconds

// Step 1: Get hours
hours = 76000 / 3600 = 21
// 21 complete hours

// Step 2: Remove hours, get remainder
remaining = 76000 % 3600 = 400
// 400 seconds left after removing 21 hours

// Step 3: Get minutes from remainder
minutes = 400 / 60 = 6
// 6 complete minutes

// Step 4: Get final seconds
seconds = 400 % 60 = 40
// 40 seconds remain

// Output: 21:HOUR 06:MIN :40SEC ✓

// Verification:
// (21 × 3600) + (6 × 60) + 40 = 75600 + 360 + 40 = 76000 ✓

5️⃣ Understanding setfill and setw

Formatting the Output with Leading Zeros:

📌 Using iomanip for Formatting:

#include <iomanip>

// setfill('0') - Fill empty spaces with '0'
// setw(2) - Set width to 2 characters

cout << setfill('0') << setw(2) << 5;  // Prints: "05"
cout << setfill('0') << setw(2) << 15; // Prints: "15"

// Why we need this:
// Without formatting: "15:HOUR 5:MIN :0SEC" (looks wrong!)
// With formatting:    "15:HOUR 05:MIN :00SEC" (perfect!)

Manipulator	Purpose	Example
`setfill('0')`	Set the fill character	Fills empty spaces with '0'
`setw(2)`	Set minimum width	Ensures at least 2 digits

6️⃣ Complete Solution Breakdown

📌 Full Code Explanation:

#include <iostream>
#include <iomanip>

using namespace std;

int main() {
    long long sec;
    cin >> sec;
    
    // Extract hours (3600 seconds = 1 hour)
    long long hours = sec / 3600;
    sec = sec % 3600;  // Update sec to remainder
    
    // Extract minutes (60 seconds = 1 minute)
    long long minutes = sec / 60;
    sec = sec % 60;    // Final seconds
    
    // Format and print with leading zeros
    cout << setfill('0') << setw(2) << hours << ":HOUR "
         << setw(2) << minutes << ":MIN :"
         << setw(2) << sec << "SEC";
    
    return 0;
}

7️⃣ Real-World Applications

This technique is used everywhere:

Digital Clocks: Converting timestamps to readable time
Video Players: Showing duration (2:34:15)
Fitness Apps: Workout time tracking
Gaming: Countdown timers, session duration
Currency Conversion: Breaking down dollars into bills and coins
Date Calculations: Days → Years, Months, Days
File Size Display: Bytes → GB, MB, KB

8️⃣ Extension Challenge: Other Conversions

📌 Same Logic, Different Units:

// Convert cents to dollars and cents
int totalCents = 567;
int dollars = totalCents / 100;  // 5 dollars
int cents = totalCents % 100;     // 67 cents

// Convert days to years, months, days (simplified)
int totalDays = 400;
int years = totalDays / 365;      // 1 year
int days = totalDays % 365;       // 35 days

// Convert bytes to GB, MB, KB
long long bytes = 5368709120;  // 5 GB
long long gb = bytes / (1024*1024*1024);
bytes = bytes % (1024*1024*1024);
long long mb = bytes / (1024*1024);

                            🎓 Key Takeaway: The combination of / and % 
                            is a fundamental pattern for decomposing values into hierarchical units. Master this, 
                            and you can convert anything!
                        

💡 Pro Tip: The Mathematical Invariant

For any division a / b and a % b:

a = (a / b) × b + (a % b)

This means you can always reconstruct the original number from the quotient and remainder!
Example: 54300 = (15 × 3600) + (5 × 60) + 0 ✓

💻 Solution Code

#include <iostream>
#include <iomanip>

using namespace std;

int main() {
    long long sec;
    cin >> sec;
    
    long long hours = sec / 3600;
    sec = sec % 3600;
    
    long long minutes = sec / 60;
    sec = sec % 60;
    
    cout << setfill('0') << setw(2) << hours << ":HOUR "
         << setw(2) << minutes << ":MIN :"
         << setw(2) << sec << "SEC";
    
    return 0;
}