Sync LeetCode submission Runtime - 10 ms (98.44%), Memory - 25.1 MB (68.75%)

Author: jimit105

2090-number-of-ways-to-arrive-at-destination/README.md

@@ -0,0 +1,41 @@
+<p>You are in a city that consists of <code>n</code> intersections numbered from <code>0</code> to <code>n - 1</code> with <strong>bi-directional</strong> roads between some intersections. The inputs are generated such that you can reach any intersection from any other intersection and that there is at most one road between any two intersections.</p>
+
+<p>You are given an integer <code>n</code> and a 2D integer array <code>roads</code> where <code>roads[i] = [u<sub>i</sub>, v<sub>i</sub>, time<sub>i</sub>]</code> means that there is a road between intersections <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> that takes <code>time<sub>i</sub></code> minutes to travel. You want to know in how many ways you can travel from intersection <code>0</code> to intersection <code>n - 1</code> in the <strong>shortest amount of time</strong>.</p>
+
+<p>Return <em>the <strong>number of ways</strong> you can arrive at your destination in the <strong>shortest amount of time</strong></em>. Since the answer may be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2025/02/14/1976_corrected.png" style="width: 255px; height: 400px;" />
+<pre>
+<strong>Input:</strong> n = 7, roads = [[0,6,7],[0,1,2],[1,2,3],[1,3,3],[6,3,3],[3,5,1],[6,5,1],[2,5,1],[0,4,5],[4,6,2]]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> The shortest amount of time it takes to go from intersection 0 to intersection 6 is 7 minutes.
+The four ways to get there in 7 minutes are:
+- 0 ➝ 6
+- 0 ➝ 4 ➝ 6
+- 0 ➝ 1 ➝ 2 ➝ 5 ➝ 6
+- 0 ➝ 1 ➝ 3 ➝ 5 ➝ 6
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 2, roads = [[1,0,10]]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> There is only one way to go from intersection 0 to intersection 1, and it takes 10 minutes.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 200</code></li>
+	<li><code>n - 1 &lt;= roads.length &lt;= n * (n - 1) / 2</code></li>
+	<li><code>roads[i].length == 3</code></li>
+	<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n - 1</code></li>
+	<li><code>1 &lt;= time<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
+	<li><code>u<sub>i </sub>!= v<sub>i</sub></code></li>
+	<li>There is at most one road connecting any two intersections.</li>
+	<li>You can reach any intersection from any other intersection.</li>
+</ul>

2090-number-of-ways-to-arrive-at-destination/solution.py

@@ -0,0 +1,43 @@
+# Approach 1: Dijkstra's Algorithm
+
+# e = total no. of edges
+# Time: O((n + e) log n)
+# Space: O(n + e)
+
+import heapq
+
+class Solution:
+    def countPaths(self, n: int, roads: List[List[int]]) -> int:
+        MOD = 10 ** 9 + 7
+
+        graph = [[] for _ in range(n)]
+
+        for start_node, end_node, travel_time in roads:
+            graph[start_node].append((end_node, travel_time))
+            graph[end_node].append((start_node, travel_time))
+
+        min_heap = [(0, 0)] # (time, node)
+        shortest_time = [float('inf')] * n
+        path_count = [0] * n # No. of ways to reach node in shortest time
+
+        shortest_time[0] = 0
+        path_count[0] = 1
+
+        while min_heap:
+            curr_time, curr_node = heapq.heappop(min_heap)
+            if curr_time > shortest_time[curr_node]:
+                continue
+
+            for neighbor_node, road_time in graph[curr_node]:
+                # Found a new shortest path → Update shortest time and reset path count
+                if curr_time + road_time < shortest_time[neighbor_node]:
+                    shortest_time[neighbor_node] = curr_time + road_time
+                    path_count[neighbor_node] = path_count[curr_node]
+                    heapq.heappush(min_heap, (shortest_time[neighbor_node], neighbor_node))
+
+                # Found another way with the same shortest time → Add to path count
+                elif curr_time + road_time == shortest_time[neighbor_node]:
+                    path_count[neighbor_node] = (path_count[neighbor_node] + path_count[curr_node]) % MOD
+
+        return path_count[n - 1]
+