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

        // rows contained
        map<int, int> count;

        vector<int> res(2);
        int best = 1e9;

        int b = 0;
        for (int k = 0; k < v.size(); k++){
            const auto& [val, i, j] = v[k];

            // add elem
            count[i]++;

            // check validity
            if (count.size() == nums.size()){
                // try to slide back up
                while (b < v.size() && count[get<1>(v[b])] > 1){
                    count[get<1>(v[b])]--;
                    b++;
                }

                // update best 
                const int back_val = get<0>(v[b]);
                const int range = val - back_val + 1;
                
                if (range < best){
                    best = range;
                    res = {back_val, val};
                }
            }
        }

        return res;
    }
};
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