<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 r = 0;
        int maxi = 0;
        int n = nums.size();
        vector<int> temp;
        for (auto& i : nums)
            for (auto& j : i)
                temp.push_back(j);
        sort(temp.begin(), temp.end());

        temp.resize(unique(temp.begin(), temp.end()) - temp.begin());

        for (auto& i : nums)
            for (auto& j : i)
                maxi = max(maxi, j), r = min(r, j);

        r++;
        pair<int, int> ans = {0, 1e9};
        auto check = [&](int l, int r) {
            for (auto& i : nums) {
                auto it = lower_bound(i.begin(), i.end(), l);
                if (it == i.end())
                    return false;
                else if (*it > r)
                    return false;
            }
            return true;
        };

        int ind = 0;
        while (r <= maxi && ind < temp.size()) {
            int l = temp[ind++];
            // use binary search
            {
                int low = r, high = maxi + 1;
                while (low <= high) {
                    int mid = (low + high) >> 1ll;

                    if (check(l, mid)) {
                        r = mid;
                        high = mid - 1;
                    } else {
                        low = mid + 1;
                    }
                }
            }

            while (r <= maxi && !check(l, r))
                r++;

            if (check(l, r)) {
                if (r - l < ans.second - ans.first) {
                    ans = make_pair(l, r);
                }
            } else
                r++;
        }
        vector<int> res = {ans.first, ans.second};
        return res;
    }
};
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