How to Solve Cheapest Flights Within K Stops Leetcode Problem - Interview Coder Guide
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How to Solve Cheapest Flights Within K Stops Problem

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Medium#803
LeetCode Problem

Cheapest Flights Within K Stops

There are n cities connected by some number of flights. You are given an array flights where flights[i] = [fromi, toi, pricei] indicates that there is a flight from city fromi to city toi with cost pr...

Dynamic ProgrammingDepth-First SearchBreadth-First Search+3 more

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Problem Breakdown

Understanding the Cheapest Flights Within K Stops Problem

Let's break down this LeetCode problem and understand what makes it challenging in interview settings.

Problem Statement

There are n cities connected by some number of flights. You are given an array flights where flights[i] = [fromi, toi, pricei] indicates that there is a flight from city fromi to city toi with cost pricei. You are also given three integers src, dst, and k, return the cheapest price from src to dst with at most k stops. If there is no such route, return -1.

MediumProblem #803
LeetCode

Cheapest Flights Within K Stops

Related Topics
Dynamic ProgrammingDepth-First SearchBreadth-First SearchGraphHeap (Priority Queue)Shortest Path
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Examples

Example 1
INPUT
n = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1
OUTPUT
700
EXPLANATION

The graph is shown above. The optimal path with at most 1 stop from city 0 to 3 is marked in red and has cost 100 + 600 = 700. Note that the path through cities [0,1,2,3] is cheaper but is invalid because it uses 2 stops.

Example 2
INPUT
n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1
OUTPUT
200
EXPLANATION

The graph is shown above. The optimal path with at most 1 stop from city 0 to 2 is marked in red and has cost 100 + 100 = 200.

Example 3
INPUT
n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0
OUTPUT
500
EXPLANATION

The graph is shown above. The optimal path with no stops from city 0 to 2 is marked in red and has cost 500.

Constraints

1 <= n <= 100
0 <= flights.length <= (n * (n - 1) / 2)
flights[i].length == 3
0 <= fromi, toi < n
fromi != toi
1 <= pricei <= 104
There will not be any multiple flights between two cities.
0 <= src, dst, k < n
src != dst

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