<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) {
        // brute force 
        int sm = INT_MAX, lar = INT_MIN, n = arrays.size();
        
        vector<int> vecsm(n,0), veclar(n,0);
        for(int i=0;i<n;i++){
            sm = INT_MAX, lar = INT_MIN;
            for(int j=0;j<arrays[i].size();j++){
                sm = min(sm, arrays[i][j]);
                lar = max(lar, arrays[i][j]);
            }
            vecsm[i] = sm;
            veclar[i] = lar;
        }
        vector<int> copy;
        copy = veclar;
        sort(copy.begin(), copy.end());
        int ans = 0;
        for(int i=0;i<n;i++){
            if(copy[0] == veclar[i]){
                ans = max(ans, abs(vecsm[i] - copy[1]));
            }else {
                ans = max(ans, abs(vecsm[i] - copy[0]));
            }
            if(copy[n-1] == veclar[i]){
                ans = max(ans, abs(vecsm[i] - copy[n-2]));
            }else{
                ans = max(ans, abs(vecsm[i] - copy[n-1]));
            }
        }
        return ans;
        
    }
};
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