Calculate Delayed Arrival Time - Solution & Explanation
Problem Statement
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 arrive at the station.
Note that the time in this problem is in 24-hours format.
Example 1:
Input: arrivalTime = 15, delayedTime = 5 Output: 20 Explanation: Arrival time of the train was 15:00 hours. It is delayed by 5 hours. Now it will reach at 15+5 = 20 (20:00 hours).
Example 2:
Input: arrivalTime = 13, delayedTime = 11 Output: 0 Explanation: Arrival time of the train was 13:00 hours. It is delayed by 11 hours. Now it will reach at 13+11=24 (Which is denoted by 00:00 in 24 hours format so return 0).
Constraints:
1 <= arrivaltime < 241 <= delayedTime <= 24
Approach Overview
Problem Overview: You are given an arrivalTime (0–23) and a delayedTime. The task is to compute the new arrival hour after the delay while keeping the result within a 24‑hour clock. If the sum exceeds 23, the clock wraps around to the beginning.
Approach 1: Simple Addition and Modulus (O(1) time, O(1) space)
The most direct solution adds the delay to the original arrival hour and applies modulo 24. The modulo operation automatically wraps values back into the valid range of a 24‑hour clock. The calculation is (arrivalTime + delayedTime) % 24. This works because modular arithmetic models cyclic systems like clocks. The approach uses constant time and space since it performs only a single arithmetic operation and returns the result.
This technique relies on basic math and modular arithmetic. Whenever a problem involves circular ranges (hours, days of week, rotating indices), modulo arithmetic provides a clean and reliable solution.
Approach 2: Alternative Calculation using Conditional Logic (O(1) time, O(1) space)
You can also compute the result using conditional checks instead of modulo. First calculate sum = arrivalTime + delayedTime. If the sum is less than 24, it is already a valid hour. If it is 24 or greater, subtract 24 to wrap the value back into the correct range. This mirrors how a physical clock resets after midnight.
This method uses straightforward arithmetic and conditional branching. It may feel more intuitive if you are thinking about the real-world behavior of clocks rather than mathematical cycles. However, the modulo version is shorter and scales better when working with other cyclic problems.
Recommended for interviews: Interviewers generally expect the modulo solution because it demonstrates comfort with cyclic calculations and produces the cleanest code. The conditional version shows you understand the underlying behavior, but the expression (arrivalTime + delayedTime) % 24 communicates the intent immediately and avoids extra branching.
Approach 1: Simple Addition and Modulus
This approach involves using simple arithmetic to add the delayed time to the arrival time and then using the modulus operator to handle the wrap-around effect that occurs after 23:00 on a 24-hour clock. By taking modulus of the sum with 24, you ensure that the resulting time stays within the 0 to 23 range, effectively resetting to 0 after 23.
In the C solution, we define a function delayedArrivalTime which takes two parameters: arrivalTime and delayedTime. The function calculates the new arrival time by adding these two values and taking the modulus with 24.
Complexity
Time Complexity: O(1) since the calculation involves basic arithmetic operations.
Space Complexity: O(1) as no extra space is used beyond the inputs provided.
Approach 2: Alternative Calculation using Conditional Logic
Instead of using the modulus operator directly, this approach utilizes conditional logic to determine if the calculated time exceeds 23:00. If it does, the result is adjusted by subtracting 24, effectively wrapping it back to start from 0.
In this C solution, an alternative function delayedArrivalTimeAlt is defined where conditional logic checks if the summed time exceeds 24 and adjusts accordingly. The conditional operator achieves this succinctly.
Complexity
Time Complexity: O(1) as it sticks to constant time operations.
Space Complexity: O(1) given no additional structures are used.
Approach 3: Default Approach
Try this approach in the editor →Complexity Comparison
| Approach | Complexity |
|---|---|
| Simple Addition and Modulus | Time Complexity: O(1) since the calculation involves basic arithmetic operations. |
| Alternative Calculation using Conditional Logic | Time Complexity: O(1) as it sticks to constant time operations. |
| Default Approach | — |
Detailed Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Simple Addition and Modulus | O(1) | O(1) | Best general solution for clock-style wraparound problems |
| Conditional Logic with Wraparound | O(1) | O(1) | Useful when modulo is avoided or when explaining the clock behavior step by step |
Video Solution
Leetcode | 2651. Calculate Delayed Arrival Time | Easy | Java Solution • Developer Docs • 672 views views
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