Create number-of-ways-to-arrive-at-destination.py

Author: kamyu104

Python/number-of-ways-to-arrive-at-destination.py

@@ -0,0 +1,43 @@
+# Time:  O((|E| + |V|) * log|V|) = O(|E| * log|V|),
+#        if we can further to use Fibonacci heap, it would be O(|E| + |V| * log|V|)
+# Space: O(|E| + |V|) = O(|E|)
+
+import heapq
+
+
+class Solution(object):
+    def countPaths(self, n, roads):
+        """
+        :type n: int
+        :type roads: List[List[int]]
+        :rtype: int
+        """
+        MOD = 10**9+7
+
+        def dijkstra(adj, start, target):
+            best = collections.defaultdict(lambda:float("inf"))
+            best[start] = 0
+            min_heap = [(0, start)]
+            dp = [0]*(len(adj))  # modified, add dp to keep number of ways
+            dp[0] = 1
+            while min_heap:
+                curr, u = heapq.heappop(min_heap)
+                if best[u] < curr:
+                    continue
+                if u == target:  # modified, early return
+                    break
+                for v, w in adj[u]:                
+                    if v in best and best[v] <= curr+w:
+                        if best[v] == curr+w:  # modified, update number of ways in this minimal time
+                            dp[v] = (dp[v]+dp[u])%MOD
+                        continue
+                    dp[v] = dp[u]  # modified, init number of ways in this minimal time
+                    best[v] = curr+w
+                    heapq.heappush(min_heap, (curr+w, v))
+            return dp[target]
+
+        adj = [[] for i in xrange(n)]
+        for u, v, w in roads:
+            adj[u].append((v, w))
+            adj[v].append((u, w))
+        return dijkstra(adj, 0, n-1)