<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) {
        int minn = INT32_MAX, maxx = INT32_MIN, result = INT32_MIN;
vector<pair<int, int>>min_max;
map<int, int>mp;
int a = arrays.size(), b;
for (int i = 0; i < a; ++i) {
    b = arrays[i].size();
    for (int j = 0; j < b; ++j) {
        mp[arrays[i][j]]++;
        minn = min(arrays[i][j], minn);
        maxx = max(arrays[i][j], maxx);
    }
    min_max.push_back({ maxx, minn });
    minn = INT32_MAX;
    maxx = INT32_MIN;
}
sort(min_max.begin(), min_max.end());
reverse(min_max.begin(), min_max.end());
int counter = min_max.size();
for (int i = 0; i < counter; ++i) {
    minn = min_max[i].second;
    if (min_max[i].first == min_max[0].first && mp[min_max[i].first] >= 2) {
        result = max(abs(min_max[i].first - minn), result);
    }
    else if (min_max[i].first != min_max[0].first) {
        result = max(abs(min_max[0].first - minn), result);
    }
    else {
        result = max(abs(min_max[1].first - minn), result);
    }
}
        return result;
    }
};
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