<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>>& a) {
        int n=a.size();
    
        vector<int> prefix_max(n);
        vector<int> suffix_max(n);
        prefix_max[0]=a[0][a[0].size()-1];
        for(int i=1;i<n;i++){
            prefix_max[i]=max(prefix_max[i-1], a[i][a[i].size()-1]);
        }
        suffix_max[n-1]=a[n-1][a[n-1].size()-1];
        for(int i=n-2;i>=0; i--){
            suffix_max[i]=max(suffix_max[i+1],a[i][a[i].size()-1]);
        }

        int maxi=0;
        for(int i=0;i<n;i++){
            int maxi_element=INT_MIN;
            if(i>0){
                maxi_element=max(maxi_element, prefix_max[i-1]);
            }
            if(i<n-1){
                maxi_element=max(maxi_element,suffix_max[i+1]);
            }
            maxi=max(maxi, maxi_element-a[i][0]);

        }
        return maxi;

    }
};
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