Cheapest Flights Within K Stops: coding question

Problem

There are n cities with the numbers 0 to n - 1. Each item [from, to, price] in flights is a one-way flight from city from to city to. The flight costs price.

Find the cheapest route from city src to city dst that has at most k stops. A stop is a city on the route between src and dst.

Return the cost of the cheapest route. If no such route exists, return -1.

Example 1

Input:: n = 4, flights = [[0,1,50],[1,3,50],[0,2,10],[2,1,10],[0,3,200]], src = 0, dst = 3, k = 1
Output:: 100
Explanation:: The route 0, 2, 1, 3 costs only 70, but it has 2 stops. The cheapest route with at most 1 stop is 0, 1, 3 for 100.

Example 2

Input:: n = 4, flights = [[0,1,50],[1,3,50],[0,2,10],[2,1,10],[0,3,200]], src = 0, dst = 3, k = 2
Output:: 70
Explanation:: Now the route 0, 2, 1, 3 has an allowed number of stops. It costs 10 + 10 + 50, which is 70.

Example 3

Input:: n = 3, flights = [[0,1,5]], src = 0, dst = 2, k = 1
Output:: -1
Explanation:: No flight goes to city 2.

Constraints

1 <= n <= 100
0 <= flights.length <= n * (n - 1) / 2
flights[i].length == 3
0 <= from, to < n
from != to
1 <= price <= 10^4
No two flights have the same from and to
0 <= src, dst, k < n
src != dst

SysRound wrote this statement. A reference solution checks every example.

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Illumio
Leetcode 787. Cheapest Flights Within K Stops
Reported Oct 1, 2026 · 1 source report
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Leetcode 787. Cheapest Flights Within K Stops
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Cheapest Flights Within K Stops
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Oracle
Cheapest Flights Within K Stops

Problem

There are n cities with the numbers 0 to n - 1. Each item [from, to, price] in flights is a one-way flight from city from to city to. The flight costs price.

Find the cheapest route from city src to city dst that has at most k stops. A stop is a city on the route between src and dst.

Return the cost of the cheapest route. If no such route exists, return -1.

Example 1

Input:: n = 4, flights = [[0,1,50],[1,3,50],[0,2,10],[2,1,10],[0,3,200]], src = 0, dst = 3, k = 1
Output:: 100
Explanation:: The route 0, 2, 1, 3 costs only 70, but it has 2 stops. The cheapest route with at most 1 stop is 0, 1, 3 for 100.

Example 2

Input:: n = 4, flights = [[0,1,50],[1,3,50],[0,2,10],[2,1,10],[0,3,200]], src = 0, dst = 3, k = 2
Output:: 70
Explanation:: Now the route 0, 2, 1, 3 has an allowed number of stops. It costs 10 + 10 + 50, which is 70.

Example 3

Input:: n = 3, flights = [[0,1,5]], src = 0, dst = 2, k = 1
Output:: -1
Explanation:: No flight goes to city 2.

Constraints

1 <= n <= 100
0 <= flights.length <= n * (n - 1) / 2
flights[i].length == 3
0 <= from, to < n
from != to
1 <= price <= 10^4
No two flights have the same from and to
0 <= src, dst, k < n
src != dst

SysRound wrote this statement. A reference solution checks every example.

Cheapest Flights Within K Stops

Problem

Example 1

Example 2

Example 3

Constraints

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Cheapest Flights Within K Stops

Problem

Example 1

Example 2

Example 3

Constraints

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