Simplest & Efficient 1-Liner C Code Time[O(1)], Space[O(1)]🔥
Intuition\nThe problem asks for finding the delayed arrival time given the original arrival time and the amount of delay. We need to consider the possibility of
# Intuition\nThe problem asks for finding the delayed arrival time given the original arrival time and the amount of delay. We need to consider the possibility of the arrival time crossing midnight (i.e., going beyond 24 hours).\n\n# Approach\nTo solve this problem, we can use the modulus operator (%), which calculates the remainder when the arrival time and delayed time are divided by 24. This operation ensures that the result is always within the range of 0 to 23, representing the hours of a day. By adding the arrival time and delayed time together and then taking the modulus 24, we can find the delayed arrival time.\n\nFor example, if the arrival time is 20 (8 PM) and the delayed time is 6, adding them together gives us 26. Taking the modulus 24 of 26, we get 2, which represents 2 AM, indicating the delayed arrival time.\n\n# Complexity\n- Time complexity: The time complexity of this approach is $O(1)$ since it only involves a simple calculation and a modulus operation, which are constant time operations.\n\n- Space complexity: The space complexity is also $O(1)$ as there is no additional space used apart from the return value.\n\n# Code\n```\nint findDelayedArrivalTime(int arrivalTime, int delayedTime){\n    return (arrivalTime + delayedTime)%24;}\n```
['C']