Time: 264 ms (78.18%), Space: 17.7 MB (60.27%) - LeetHub

Author: Deepak5113

1334-find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance/1334-find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance.py

@@ -0,0 +1,35 @@
+class Solution:
+    def findTheCity(self, n: int, edges: List[List[int]], distanceThreshold: int) -> int:
+        inf = float('inf')
+        # Step 1: Initialize the distance matrix with inf
+        dist = [[inf] * n for _ in range(n)]
+        
+        # Distance from a city to itself is 0
+        for i in range(n):
+            dist[i][i] = 0
+        
+        # Step 2: Fill the initial distances from the edges
+        for u, v, w in edges:
+            dist[u][v] = w
+            dist[v][u] = w
+        
+        # Step 3: Apply Floyd-Warshall Algorithm to find shortest paths between all pairs of cities
+        for k in range(n):
+            for i in range(n):
+                for j in range(n):
+                    if dist[i][j] > dist[i][k] + dist[k][j]:
+                        dist[i][j] = dist[i][k] + dist[k][j]
+        
+        # Step 4: Count reachable cities within distance threshold for each city
+        min_count = inf
+        result_city = -1
+        
+        for i in range(n):
+            count = sum(1 for j in range(n) if dist[i][j] <= distanceThreshold)
+            
+            # Choose the city with the smallest count, and in case of ties, the city with the greatest index
+            if count < min_count or (count == min_count and i > result_city):
+                min_count = count
+                result_city = i
+        
+        return result_city