<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>

#define minimum first
#define maximum second
#define INF 10000000
class Solution {
public:
    int maxDistance(vector<vector<int>>& arrays) {
        int n = arrays.size();

        vector<pair<int,int>> minMax(n, {INF, -INF});
        for(int i = 0; i < n; ++i){
            for(auto &num : arrays[i]){
                minMax[i].minimum = min(minMax[i].minimum, num);
                minMax[i].maximum = max(minMax[i].maximum, num);
            }
        }

        // setting suffix min and max
        vector<pair<int,int>> suff(n, {INF, -INF}); // {suffixMin, suffixMax}
        for(int i = n-2; i >= 0; --i){
            auto [nextMin, nextMax] = minMax[i+1];
            suff[i].minimum = min(suff[i+1].minimum, nextMin);
            suff[i].maximum = max(suff[i+1].maximum, nextMax);
        }

        // set answer
        int ans = 0, prefixMin = INF, prefixMax = -INF;
        for(int i = 0; i < n; ++i){
            auto [currMin, currMax] = minMax[i];
            for(auto &a : {currMin, currMax}){
                for(auto &b : {prefixMin, prefixMax, suff[i].minimum, suff[i].maximum}){
                    if(abs(b) < INF)
                        ans = max(ans, abs(a-b));
                }
            }
            prefixMin = min(prefixMin, currMin);
            prefixMax = max(prefixMax, currMax);
        }

        return ans;
    }
};
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