<p>You have <code>k</code> lists of sorted integers in <strong>non-decreasing&nbsp;order</strong>. Find the <b>smallest</b> range that includes at least one number from each of the <code>k</code> lists.</p>

<p>We define the range <code>[a, b]</code> is smaller than range <code>[c, d]</code> if <code>b - a &lt; d - c</code> <strong>or</strong> <code>a &lt; c</code> if <code>b - a == d - c</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [[4,10,15,24,26],[0,9,12,20],[5,18,22,30]]
<strong>Output:</strong> [20,24]
<strong>Explanation: </strong>
List 1: [4, 10, 15, 24,26], 24 is in range [20,24].
List 2: [0, 9, 12, 20], 20 is in range [20,24].
List 3: [5, 18, 22, 30], 22 is in range [20,24].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [[1,2,3],[1,2,3],[1,2,3]]
<strong>Output:</strong> [1,1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>nums.length == k</code></li>
	<li><code>1 &lt;= k &lt;= 3500</code></li>
	<li><code>1 &lt;= nums[i].length &lt;= 50</code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i][j] &lt;= 10<sup>5</sup></code></li>
	<li><code>nums[i]</code>&nbsp;is sorted in <strong>non-decreasing</strong> order.</li>
</ul>

class Solution {
public:
    vector<int> smallestRange(vector<vector<int>>& nums) {
        int n = nums.size();
        vector<pair<int, int>> v;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < nums[i].size(); j++) {
                v.push_back({nums[i][j], i});
            }
        }
        sort(v.begin(), v.end());

        int fcnt = 0;
        int j = 0;
        int mp[n];
        int mini = INT_MAX;
        int l = 0, r = 0;
        memset(mp, 0, sizeof(mp));

        for (int i = 0; i < v.size(); i++) {
            mp[v[i].second]++;
            if (mp[v[i].second] == 1)
                fcnt++;

            if (fcnt == n) {
                while (fcnt == n) {
                    if (mp[v[j].second] > 1) {
                        mp[v[j].second]--;
                        j++;
                    } else {
                        break;
                    }
                }
                if (v[i].first - v[j].first < mini) {
                    mini = v[i].first - v[j].first;
                    l = v[j].first;
                    r = v[i].first;
                }
            }
        }
        return {l, r};
    }
};
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