# Solving Leetcode Interviews in Seconds with AI: Minimum Time Difference


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "539" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given a list of 24-hour clock time points in "HH:MM" format, return the minimum minutes difference between any two time-points in the list.   Example 1: Input: timePoints = ["23:59","00:00"] Output: 1 Example 2: Input: timePoints = ["00:00","23:59","00:00"] Output: 0    Constraints:  2 <= timePoints.length <= 2 * 104 timePoints[i] is in the format "HH:MM".  

	# Explanation
	Here's the breakdown of the approach, complexities, and the Python code:

*   **Approach:**
    1.  Convert each time point into minutes from midnight.
    2.  Sort the list of minutes.
    3.  Calculate the minimum difference between adjacent time points, considering the wraparound (difference between the last and first time points).

*   **Complexity:**
    *   Runtime Complexity: O(n log n) due to sorting.
    *   Storage Complexity: O(n) to store the minutes.

	
	# Code
	```python
	def findMinDifference(timePoints):
    """
    Finds the minimum minutes difference between any two time-points in the list.

    Args:
        timePoints (list[str]): A list of time points in "HH:MM" format.

    Returns:
        int: The minimum minutes difference between any two time-points.
    """

    minutes = []
    for time in timePoints:
        hours, mins = map(int, time.split(':'))
        minutes.append(hours * 60 + mins)

    minutes.sort()

    min_diff = float('inf')
    for i in range(len(minutes) - 1):
        min_diff = min(min_diff, minutes[i+1] - minutes[i])

    # Check the difference between the last and first time points (wraparound)
    min_diff = min(min_diff, minutes[0] + 1440 - minutes[-1])

    return min_diff
	```