Given a list of 24-hour clock time points in <strong>&quot;HH:MM&quot;</strong> format, return <em>the minimum <b>minutes</b> difference between any two time-points in the list</em>.
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> timePoints = ["23:59","00:00"]
<strong>Output:</strong> 1
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> timePoints = ["00:00","23:59","00:00"]
<strong>Output:</strong> 0
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= timePoints.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>timePoints[i]</code> is in the format <strong>&quot;HH:MM&quot;</strong>.</li>
</ul>

class Solution {
public:
    // Helper function to convert time in "HH:MM" format to minutes
    int convertToMinutes(string time) {
        int hours = stoi(time.substr(0, 2));
        int minutes = stoi(time.substr(3, 2));
        return hours * 60 + minutes;
    }

    int findMinDifference(vector<string>& times) {
        int n = times.size();
        
        // Convert all times to minutes and store in a vector
        vector<int> minutes(n);
        for (int i = 0; i < n; i++) {
            minutes[i] = convertToMinutes(times[i]);
        }
        
        // Sort the times in minutes
        sort(minutes.begin(), minutes.end());
        
        // Initialize the minimum difference with a large value
        int minDiff = INT_MAX;
        
        // Calculate the difference between consecutive times
        for (int i = 1; i < n; i++) {
            minDiff = min(minDiff, minutes[i] - minutes[i - 1]);
        }
        
        // Handle the wrap-around case (difference between first and last times)
        int wrapAroundDiff = 1440 - (minutes[n - 1] - minutes[0]);
        minDiff = min(minDiff, wrapAroundDiff);
        
        return minDiff;
    }
};

18500