<p>You are given <code>m</code> <code>arrays</code>, where each array is sorted in <strong>ascending order</strong>.</p>

<p>You can pick up two integers from two different arrays (each array picks one) and calculate the distance. We define the distance between two integers <code>a</code> and <code>b</code> to be their absolute difference <code>|a - b|</code>.</p>

<p>Return <em>the maximum distance</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arrays = [[1,2,3],[4,5],[1,2,3]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> One way to reach the maximum distance 4 is to pick 1 in the first or third array and pick 5 in the second array.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arrays = [[1],[1]]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == arrays.length</code></li>
	<li><code>2 &lt;= m &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= arrays[i].length &lt;= 500</code></li>
	<li><code>-10<sup>4</sup> &lt;= arrays[i][j] &lt;= 10<sup>4</sup></code></li>
	<li><code>arrays[i]</code> is sorted in <strong>ascending order</strong>.</li>
	<li>There will be at most <code>10<sup>5</sup></code> integers in all the arrays.</li>
</ul>

class Solution {
public:
    int maxDistance(vector<vector<int>>& arrays) {
        int n = arrays.size();
        
        // Initialize prefix_max and suffix_max arrays
        vector<int> prefix_max(n), suffix_max(n);
        
        // Fill prefix_max
        prefix_max[0] = arrays[0].back();
        for (int i = 1; i < n; ++i) {
            prefix_max[i] = max(prefix_max[i-1], arrays[i].back());
        }
        
        // Fill suffix_max
        suffix_max[n-1] = arrays[n-1].back();
        for (int i = n-2; i >= 0; --i) {
            suffix_max[i] = max(suffix_max[i+1], arrays[i].back());
        }
        
        int max_distance = 0;
        
        // Calculate the maximum distance
        for (int i = 0; i < n; ++i) {
            if (i > 0) {
                // Compare current array's min with prefix_max of the previous arrays
                max_distance = max(max_distance, abs(arrays[i].front() - prefix_max[i-1]));
            }
            if (i < n-1) {
                // Compare current array's min with suffix_max of the next arrays
                max_distance = max(max_distance, abs(arrays[i].front() - suffix_max[i+1]));
            }
        }
        
        return max_distance;
    }
};
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