<p>You are given a positive integer <code>n</code> representing <code>n</code> cities numbered from <code>1</code> to <code>n</code>. You are also given a <strong>2D</strong> array <code>roads</code> where <code>roads[i] = [a<sub>i</sub>, b<sub>i</sub>, distance<sub>i</sub>]</code> indicates that there is a <strong>bidirectional </strong>road between cities <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> with a distance equal to <code>distance<sub>i</sub></code>. The cities graph is not necessarily connected.</p>

<p>The <strong>score</strong> of a path between two cities is defined as the <strong>minimum </strong>distance of a road in this path.</p>

<p>Return <em>the <strong>minimum </strong>possible score of a path between cities </em><code>1</code><em> and </em><code>n</code>.</p>

<p><strong>Note</strong>:</p>

<ul>
	<li>A path is a sequence of roads between two cities.</li>
	<li>It is allowed for a path to contain the same road <strong>multiple</strong> times, and you can visit cities <code>1</code> and <code>n</code> multiple times along the path.</li>
	<li>The test cases are generated such that there is <strong>at least</strong> one path between <code>1</code> and <code>n</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/10/12/graph11.png" style="width: 190px; height: 231px;" />
<pre>
<strong>Input:</strong> n = 4, roads = [[1,2,9],[2,3,6],[2,4,5],[1,4,7]]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The path from city 1 to 4 with the minimum score is: 1 -&gt; 2 -&gt; 4. The score of this path is min(9,5) = 5.
It can be shown that no other path has less score.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/10/12/graph22.png" style="width: 190px; height: 231px;" />
<pre>
<strong>Input:</strong> n = 4, roads = [[1,2,2],[1,3,4],[3,4,7]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The path from city 1 to 4 with the minimum score is: 1 -&gt; 2 -&gt; 1 -&gt; 3 -&gt; 4. The score of this path is min(2,2,4,7) = 2.
</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= roads.length &lt;= 10<sup>5</sup></code></li>
	<li><code>roads[i].length == 3</code></li>
	<li><code>1 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li><code>1 &lt;= distance<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
	<li>There are no repeated edges.</li>
	<li>There is at least one path between <code>1</code> and <code>n</code>.</li>
</ul>

class DisjointSet {
    vector<int> size, parent, minDistance;
public:
    DisjointSet(int n) {
        size.resize(n+1, 1);
        parent.resize(n+1);
        minDistance.resize(n+1, INT_MAX);
        for(int i = 0; i <= n; i++) {
            parent[i] = i;
        }
    }

    int getMinDistance(int x) {
        return minDistance[x];
    }

    int getUltimateParent(int x) {
        if(x == parent[x])
            return x;
        // path compression
        parent[x] = getUltimateParent(parent[x]);
        return parent[x];
    }

    void unionBySize(int a, int b, int d) {
        int ulp_a = getUltimateParent(a);
        int ulp_b = getUltimateParent(b);

        if(ulp_a == ulp_b) {
            minDistance[ulp_a] = min(minDistance[ulp_a], d);
            return;
        }

        if(size[ulp_a] < size[ulp_b]) {
            parent[ulp_a] = ulp_b;
            size[ulp_b] += size[ulp_a];
            minDistance[ulp_b] = min(d, min(minDistance[ulp_a], minDistance[ulp_b]));
        } else {
            parent[ulp_b] = ulp_a;
            size[ulp_a] += size[ulp_b];
            minDistance[ulp_a] = min(d, min(minDistance[ulp_a], minDistance[ulp_b]));
        }
    }
};



class Solution {
public:
    int minScore(int n, vector<vector<int>>& roads) {
        DisjointSet ds(n);

        for(auto road: roads) {
            int a = road[0], b = road[1], d = road[2];
            ds.unionBySize(a, b, d);
        }

        return ds.getMinDistance(ds.getUltimateParent(1));
    }
};
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