<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>

typedef pair<int, int> pi;
class Solution {
public:
    int maxDistance(vector<vector<int>>& arrays) {
        int n = arrays.size();
        vector<int> mn(n, INT_MAX), mx(n, INT_MIN);
        for (int i = 0; i < n; ++i)
            for (auto& v : arrays[i]) 
                mn[i] = min(mn[i], v), mx[i] = max(mx[i], v);
        priority_queue<pi> pq1;
        priority_queue<pi, vector<pi>, greater<pi>> pq2;
        for (int i = 0; i < n; ++i)
            pq1.push({mx[i], i}), pq2.push({mn[i], i});
        auto mx1 = pq1.top(); pq1.pop();
        auto mx2 = pq1.top();
        auto mn1 = pq2.top(); pq2.pop();
        auto mn2 = pq2.top();
        pi mxs[2] = {mx1, mx2};
        pi mns[2] = {mn1, mn2};
        int ans = 0;
        for (auto& u : mxs) {
            for (auto& v : mns) {
                if (u.second != v.second) {
                    ans = max(ans, u.first - v.first);
                }
            }
        }
        return ans;
    }
};
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