Calculate Delayed Arrival Time

Calculate Delayed Arrival Time - Problem

Math Easy

You are given a positive integer arrivalTime denoting the arrival time of a train in hours, and another positive integer delayedTime denoting the amount of delay in hours.

Return the time when the train will actually arrive at the station.

Note that the time in this problem is in 24-hours format.

Input & Output

Example 1 — Basic Case

$ Input: arrivalTime = 15, delayedTime = 5

› Output: 20

💡 Note: Train was scheduled at 15:00, delayed by 5 hours, so arrives at 20:00

Example 2 — Wraparound Case

$ Input: arrivalTime = 22, delayedTime = 5

› Output: 3

💡 Note: 22 + 5 = 27, but 27 % 24 = 3, so train arrives at 03:00 next day

Example 3 — Large Delay

$ Input: arrivalTime = 10, delayedTime = 50

› Output: 12

💡 Note: 10 + 50 = 60, and 60 % 24 = 12, so arrives at 12:00 after multiple day cycles

Constraints

1 ≤ arrivalTime ≤ 23
1 ≤ delayedTime ≤ 10⁹

Visualization

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Asked in

M Meta 15 G Google 12

The key insight is recognizing this as modular arithmetic - the 24-hour clock cycles back to 0 after 23. Best approach uses (arrivalTime + delayedTime) % 24 for instant calculation. Time: O(1), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(delayedTime) Space: O(1)

Simulate adding each hour of delay individually. Every time we reach hour 24, we wrap around to hour 0. Continue until all delay hours are added.

Modulo Arithmetic

⏱️ Time: O(1) Space: O(1)

Add the total delay to arrival time, then use modulo 24 to handle wraparound. This leverages the mathematical property that clock arithmetic is modular arithmetic.

Brute Force — Algorithm Steps

Start with original arrival time
Add delay hours one by one
Wrap to 0 whenever time reaches 24

Visualization

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Step-by-Step Walkthrough

Start

Begin with arrival time

Add Hours

Add each delay hour individually

Wrap Around

Reset to 0 when reaching 24

Code -

solution.c — C

#include <stdio.h>

int solution(int arrivalTime, int delayedTime) {
    int currentTime = arrivalTime;
    for (int i = 0; i < delayedTime; i++) {
        currentTime += 1;
        if (currentTime == 24) {
            currentTime = 0;
        }
    }
    return currentTime;
}

int main() {
    int arrivalTime, delayedTime;
    scanf("%d", &arrivalTime);
    scanf("%d", &delayedTime);
    int result = solution(arrivalTime, delayedTime);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(delayedTime)

We iterate through each hour of delay individually

✓ Linear Growth

Space Complexity

O(1)

Only using a single variable to track current time

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

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Code -

Time & Space Complexity

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