leetcode-problemset/leetcode/originData/minimum-cost-walk-in-weighted-graph.json

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            "title": "Minimum Cost Walk in Weighted Graph",
            "titleSlug": "minimum-cost-walk-in-weighted-graph",
            "content": "<p>There is an undirected weighted graph with <code>n</code> vertices labeled from <code>0</code> to <code>n - 1</code>.</p>\n\n<p>You are given the integer <code>n</code> and an array <code>edges</code>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>]</code> indicates that there is an edge between vertices <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> with a weight of <code>w<sub>i</sub></code>.</p>\n\n<p>A walk on a graph is a sequence of vertices and edges. The walk starts and ends with a vertex, and each edge connects the vertex that comes before it and the vertex that comes after it. It&#39;s important to note that a walk may visit the same edge or vertex more than once.</p>\n\n<p>The <strong>cost</strong> of a walk starting at node <code>u</code> and ending at node <code>v</code> is defined as the bitwise <code>AND</code> of the weights of the edges traversed during the walk. In other words, if the sequence of edge weights encountered during the walk is <code>w<sub>0</sub>, w<sub>1</sub>, w<sub>2</sub>, ..., w<sub>k</sub></code>, then the cost is calculated as <code>w<sub>0</sub> &amp; w<sub>1</sub> &amp; w<sub>2</sub> &amp; ... &amp; w<sub>k</sub></code>, where <code>&amp;</code> denotes the bitwise <code>AND</code> operator.</p>\n\n<p>You are also given a 2D array <code>query</code>, where <code>query[i] = [s<sub>i</sub>, t<sub>i</sub>]</code>. For each query, you need to find the minimum cost of the walk starting at vertex <code>s<sub>i</sub></code> and ending at vertex <code>t<sub>i</sub></code>. If there exists no such walk, the answer is <code>-1</code>.</p>\n\n<p>Return <em>the array </em><code>answer</code><em>, where </em><code>answer[i]</code><em> denotes the <strong>minimum</strong> cost of a walk for query </em><code>i</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">n = 5, edges = [[0,1,7],[1,3,7],[1,2,1]], query = [[0,3],[3,4]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[1,-1]</span></p>\n\n<p><strong>Explanation:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/01/31/q4_example1-1.png\" style=\"padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 351px; height: 141px;\" />\n<p>To achieve the cost of 1 in the first query, we need to move on the following edges: <code>0-&gt;1</code> (weight 7), <code>1-&gt;2</code> (weight 1), <code>2-&gt;1</code> (weight 1), <code>1-&gt;3</code> (weight 7).</p>\n\n<p>In the second query, there is no walk between nodes 3 and 4, so the answer is -1.</p>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n</div>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">n = 3, edges = [[0,2,7],[0,1,15],[1,2,6],[1,2,1]], query = [[1,2]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[0]</span></p>\n\n<p><strong>Explanation:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/01/31/q4_example2e.png\" style=\"padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 211px; height: 181px;\" />\n<p>To achieve the cost of 0 in the first query, we need to move on the following edges: <code>1-&gt;2</code> (weight 1), <code>2-&gt;1</code> (weight 6), <code>1-&gt;2</code> (weight 1).</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= edges.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>edges[i].length == 3</code></li>\n\t<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n - 1</code></li>\n\t<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>\n\t<li><code>0 &lt;= w<sub>i</sub> &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= query.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>query[i].length == 2</code></li>\n\t<li><code>0 &lt;= s<sub>i</sub>, t<sub>i</sub> &lt;= n - 1</code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    vector<int> minimumCost(int n, vector<vector<int>>& edges, vector<vector<int>>& query) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def minimumCost(self, n, edges, query):\n        \"\"\"\n        :type n: int\n        :type edges: List[List[int]]\n        :type query: List[List[int]]\n        :rtype: List[int]\n        \"\"\"\n        ",
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                    "code": "/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* minimumCost(int n, int** edges, int edgesSize, int* edgesColSize, int** query, int querySize, int* queryColSize, int* returnSize) {\n    \n}",
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                    "code": "/**\n * @param {number} n\n * @param {number[][]} edges\n * @param {number[][]} query\n * @return {number[]}\n */\nvar minimumCost = function(n, edges, query) {\n    \n};",
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                "The intended solution uses Disjoint Set Union.",
                "Notice that, if <code>u</code> and <code>v</code> are not connected then the answer is <code>-1</code>, otherwise we can use all the edges from the connected component where both belong to."
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