<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 minInd = 0;
        int minVec = 0;
        int beg = nums[0][0];
        int end = nums[0][0];
        vector<pair<int, int>> window;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i][0] < beg) {
                beg = nums[i][0];
                minVec = i;
            }
            end = max(end, nums[i][0]);
            window.push_back({nums[i][0], 0});
        }
        vector<int> ans = {beg, end};
        while (window[minVec].second < nums[minVec].size()) {
            // find new range
            beg = window[0].first; end = window[0].first;
            minVec = 0;
            for (int i = 0; i < nums.size(); i++) {
                if (window[i].first < beg) {
                    beg = window[i].first;
                    minVec = i;
                }
                end = max(end, window[i].first);
            }
            if (end - beg < ans[1] - ans[0]) {
                ans[0] = beg; ans[1] = end;
            }
            // update window
            window[minVec].second++;
            if (window[minVec].second == nums[minVec].size()) break;
            window[minVec].first = nums[minVec][window[minVec].second];
        }
        return ans;
    }
};
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