# Cheapest Flights With K Stops

May 7, 2020 by Kevin Nasto

## Description

There are n cities connected by m flights. Each flight starts from city u and arrives at v with a price w.

Now given all the cities and flights, together with starting city src and the destination dst, your task is to find the cheapest price from src to dst with up to K stops. If there is no such route, output `-1.`

``````Example 1:
Input:
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 1
Output: 200
``````
``````Example 2:
Input:
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 0
Output: 500
``````

## Solution 1 Recursive With Memoization

``````from collections import defaultdict

def get_node(name, value):
return {'id':name, 'cost':value}

def recurse(node_id, tree, k, dest, visited):

key = (node_id,k)
if key in visited:
return visited[key]

if node_id == dest:
return 0

if k < 0:
return float('inf')

children = tree[node_id]

if not children:
return float('inf')

results = []
for child in tree[node_id]:
results.append(child['cost'] + recurse(child['id'], tree, k-1, dest, visited))

visited[key] = min(results)
return visited[key]

def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, K: int) -> int:
tree = defaultdict(list)
for edge in flights:
node_id = edge[0]
child = get_node(edge[1], edge[2])
tree[node_id].append(child)

visited = {}
result = recurse(src,tree, K, dst, visited)
if result == float('inf'):
return -1
else:
return result
``````