Sync LeetCode submission Runtime - 80 ms (69.66%), Memory - 44.4 MB (68.82%)

Author: koralkulacoglu

LeetCode/0744-network-delay-time/README.md

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+<p>You are given a network of <code>n</code> nodes, labeled from <code>1</code> to <code>n</code>. You are also given <code>times</code>, a list of travel times as directed edges <code>times[i] = (u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>)</code>, where <code>u<sub>i</sub></code> is the source node, <code>v<sub>i</sub></code> is the target node, and <code>w<sub>i</sub></code> is the time it takes for a signal to travel from source to target.</p>
+
+<p>We will send a signal from a given node <code>k</code>. Return <em>the <strong>minimum</strong> time it takes for all the</em> <code>n</code> <em>nodes to receive the signal</em>. If it is impossible for all the <code>n</code> nodes to receive the signal, return <code>-1</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2019/05/23/931_example_1.png" style="width: 217px; height: 239px;" />
+<pre>
+<strong>Input:</strong> times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
+<strong>Output:</strong> 2
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> times = [[1,2,1]], n = 2, k = 1
+<strong>Output:</strong> 1
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> times = [[1,2,1]], n = 2, k = 2
+<strong>Output:</strong> -1
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= k &lt;= n &lt;= 100</code></li>
+	<li><code>1 &lt;= times.length &lt;= 6000</code></li>
+	<li><code>times[i].length == 3</code></li>
+	<li><code>1 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n</code></li>
+	<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
+	<li><code>0 &lt;= w<sub>i</sub> &lt;= 100</code></li>
+	<li>All the pairs <code>(u<sub>i</sub>, v<sub>i</sub>)</code> are <strong>unique</strong>. (i.e., no multiple edges.)</li>
+</ul>

LeetCode/0744-network-delay-time/solution.cpp

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+class Solution {
+public:
+    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
+        vector<vector<pair<int, int>>> graph(n);
+        for (auto& t : times) graph[t[0]-1].push_back({t[1]-1, t[2]});
+
+        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
+        vector<int> costs(n, 1e9);
+        costs[k-1] = 0;
+        pq.push({0, k-1});
+        while (!pq.empty()) {
+            auto [cost, node] = pq.top();
+            pq.pop();
+
+            if (cost > costs[node]) continue;
+
+            for (auto& neigh : graph[node]) {
+                int newNode = neigh.first;
+                int newCost = cost + neigh.second;
+
+                if (newCost < costs[newNode]) {
+                    costs[newNode] = newCost;
+                    pq.push({newCost, newNode});
+                }
+            }
+        }
+
+        int ans = *max_element(costs.begin(), costs.end());
+
+        return (ans < 1e9 ? ans : -1);
+    }
+};