539. Minimum Time Difference - Detailed Explanation

Problem Statement

Description:
You are given a list of time points in 24-hour clock format "HH:MM". Your task is to return the minimum difference in minutes between any two time points in the list. Note that the time difference can wrap around midnight. For example, the difference between "23:59" and "00:00" is 1 minute.

Examples:

1. Example 1

   • Input: timePoints = ["23:59", "00:00"]
   • Output: 1
   • Explanation: The two time points are 1 minute apart considering the wrap-around at midnight.

2. Example 2

   • Input: timePoints = ["01:01", "02:01", "03:00"]
   • Output: 59
   • Explanation: The minimum difference is between "02:01" and "03:00" which is 59 minutes.

3. Example 3

   • Input: timePoints = ["05:31", "22:08", "00:35"]
   • Output: 147
   • Explanation: After converting all times to minutes, the smallest difference is found when considering the circular nature of the clock.

Constraints:

• 2 ≤ timePoints.length ≤ 2 * 10^4
• Each time point is in the format "HH:MM".
• The answer is always non-negative.

Hints to Approach the Problem

1. Convert to Minutes:
   Convert each time string into minutes (0 to 1439). For instance, "HH:MM" converts to hours * 60 + minutes.

2. Sort the Times:
   Once converted, sort the list of minutes. Sorting helps in easily comparing consecutive time points.

3. Account for Wrap-Around:
   After comparing adjacent pairs, also compare the first and last time points considering the circular nature (i.e., adding 1440 minutes to the first time before comparing to the last).

4. Brute Force vs. Optimal:
   A brute-force approach could compare every pair, but sorting reduces the comparisons drastically to O(n log n) for sorting and O(n) for scanning.

Approaches

Approach 1: Brute Force (Pairwise Comparison)

Explanation:

• Idea:
  Compare every pair of time points, converting them to minutes, and compute their difference considering the wrap-around.
• Why It Falls Short:
  Although conceptually simple, this approach has a time complexity of O(n²) which is too slow for the maximum constraints (up to 20,000 time points).

Pseudocode:

for i from 0 to len(timePoints)-1:
    for j from i+1 to len(timePoints):
        diff = absoluteDifference(timePoints[i], timePoints[j])
        minDiff = min(minDiff, diff)
return minDiff

Note: The absoluteDifference function must account for the circular difference.

Approach 2: Sorting & Linear Scan (Optimal)

Explanation:

• Step 1: Convert Time to Minutes
  For each time string, compute the minutes: int(hours)*60 + int(minutes).

• Step 2: Sort the Times
  Sorting the list of minutes (values range from 0 to 1439) makes it easy to compare consecutive time points.

• Step 3: Compute Differences

  • Iterate through the sorted list and compute the difference between consecutive elements.
  • Additionally, calculate the difference between the last time and the first time plus 1440 (to account for the wrap-around).

• Step 4: Return the Minimum Difference
  The smallest computed difference is the answer.

Visual Example:

Consider timePoints = ["23:59", "00:00", "00:30"]:

• Convert to minutes: [1439, 0, 30]
• Sort: [0, 30, 1439]
• Compute differences:
  • Between 0 and 30: 30 minutes
  • Between 30 and 1439: 1409 minutes
  • Wrap-around: (0 + 1440) - 1439 = 1 minute
• The minimum difference is 1 minute.

Pseudocode:

function findMinDifference(timePoints):
    minutesList = []
    for time in timePoints:
        hours, minutes = split time by ':'
        minutesList.append(hours * 60 + minutes)
    
    sort(minutesList)
    
    minDiff = infinity
    for i from 1 to len(minutesList)-1:
        minDiff = min(minDiff, minutesList[i] - minutesList[i-1])
    
    # Check the difference between last and first (wrap-around)
    minDiff = min(minDiff, (minutesList[0] + 1440) - minutesList[-1])
    
    return minDiff