<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();
        vector<int> v, w;
        for(int i = 0; i < n; i++)
        {
            v.push_back(arrays[i][0]);
            w.push_back(arrays[i].back());
        }
        vector<int> pref_mn(n), suff_mn(n), pref_mx(n), suff_mx(n);
        pref_mn[0] = v[0];
        pref_mx[0] = w[0];
        for(int i = 1; i < n; i++) pref_mn[i] = min(pref_mn[i - 1], v[i]), pref_mx[i] = max(pref_mx[i - 1], w[i]);
        suff_mn[n - 1] = v[n - 1];
        suff_mx[n - 1] = w[n - 1];
        for(int i = n - 2; i >= 0; i--) suff_mn[i] = min(suff_mn[i + 1], v[i]), suff_mx[i] = max(suff_mx[i + 1], w[i]);
        int ans = max(abs(v[n - 1] - w[0]), abs(w[n - 1] - v[0]));
        for(int i = 1; i < n - 1; i++) 
        {
            ans = max(ans, w[i] - min(pref_mn[i - 1], suff_mn[i + 1]));
            ans = max(ans, max(pref_mx[i - 1], suff_mx[i + 1]) - v[i]);
        }
        return ans;
    }
};
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