<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>>& v) {
        map<int,int> m;
        vector<int> temp;
        for(int i=0; i<v.size(); i++)
        {
            m[v[i][0]]++;
            temp.push_back(v[i][v[i].size()-1]);
            // cout<<v[i][v[i].size()-1]<<" ";
        }
        // cout<<endl;

        int ans = 0;
        vector<pair<int,int>> vv;
        int cnt=0;
        for(auto g : m)
        {
            // cout<<g.first<<" "<<g.second<<endl;
            vv.push_back({g.first,g.second});
            cnt++;
            if(cnt==2)
            break;
        }
        for(int i=0; i<temp.size(); i++)
        {
            int mini = v[i][0];
            if(mini==vv[0].first)
            {
                if(vv[0].second>1)
                ans=max(ans,abs(mini-temp[i]));
                else
                ans=max(ans,abs(vv[1].first-temp[i]));
            }
            else
            ans=max(ans,abs(vv[0].first-temp[i]));
        }

        return ans;
    }
};
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