<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>

#include <vector>
#include <algorithm>
#include <limits>

using namespace std;

class Solution {
public:
    vector<int> smallestRange(vector<vector<int>>& nums) {
        int k = nums.size();
        vector<pair<int, int>> v; 
        
        for (int i = 0; i < k; ++i) {
            for (int num : nums[i]) {
                v.push_back({num, i});
            }
        }
        
        sort(v.begin(), v.end()); 
        
        vector<int> feq(k, 0);
        int unique_count = 0; 
        
        int left = 0, right = 0;
        int min_range = numeric_limits<int>::max();
        vector<int> result = {0, 0};
        
        while (right < v.size()) {
            if (feq[v[right].second]++ == 0) {
                ++unique_count;
            }
            while (unique_count == k) {
                if (v[right].first - v[left].first < min_range) {
                    min_range = v[right].first - v[left].first;
                    result = {v[left].first, v[right].first};
                }
                
              
                if (--feq[v[left].second] == 0) {
                    --unique_count;
                }
                ++left;
            }
            
            ++right;
        }
        
        return result;
    }
};

21646