What is the time complexity of Cheapest Flights Within K Stops?

The optimal solution has a time complexity of O(K * E).

What data structures are used to solve Cheapest Flights Within K Stops?

This problem uses Dynamic Programming, Depth-First Search, Breadth-First Search, Graph, Heap (Priority Queue), Shortest Path. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Cheapest Flights Within K Stops?

Insight 1: The core idea is to relax the edges of the graph representing the flight network `k+1` times. This relaxation process involves iterating through all flights and updating the minimum cost to reach the destination of each flight, considering the number of stops allowed. Insight 2: The use of dynamic programming with a temporary array to store the updated prices after each iteration is essential. This prevents updating a price based on a price that has already been updated in the same iteration, which would violate the stop constraint.

Which companies ask LeetCode 787. Cheapest Flights Within K Stops in interviews?

This problem is frequently asked at Airbnb, Cloudera, Coupang, DE Shaw, MakeMyTrip and 3 more companies.

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Cheapest Flights Within K Stops - LeetCode 787 Solution

Cheapest Flights Within K Stops - Complete Solution Guide

Cheapest Flights Within K Stops is LeetCode problem 787, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem asks us to find the cheapest price to travel from a source city `src` to a destination city `dst` within a maximum of `k` stops, given a list of flights represented as `flights[i] = [from_i, to_i, price_i]`. We need to return the minimum cost, or -1 if no such route exists within the given constraints. The number of cities is `n`. Crucially, a 'stop' represents visiting an intermediate city. A direct flight from `src` to `dst` involves zero stops.

Solution Approach

The provided solution uses a dynamic programming approach based on the Bellman-Ford algorithm to find the shortest path with a limited number of edges (stops). We maintain an array `prices` that stores the minimum cost to reach each city from the source city. We iterate `k+1` times, each time representing one more allowed stop. In each iteration, we update the `prices` array by considering all possible flights. A crucial part is to create a temporary copy of the `prices` array (`temp_prices`) before updating, to ensure that we only consider paths with at most `i` stops in the `i`-th iteration. This ensures correctness.

Step-by-Step Algorithm

Step 1: Initialize an array `prices` of size `n` with infinity (or MAX_VALUE) for all cities except the source city, which is initialized to 0.
Step 2: Iterate `k+1` times (representing the maximum number of stops allowed).
Step 3: In each iteration, create a temporary copy `temp_prices` of the `prices` array. This is critical to avoid incorrect updates during the iteration.
Step 4: Iterate through the `flights` array. For each flight `[u, v, p]`, check if `prices[u]` is not infinity. If it's not, check if `prices[u] + p` (the cost to reach city `u` plus the cost of the flight from `u` to `v`) is less than `temp_prices[v]` (the current minimum cost to reach city `v`).
Step 5: If `prices[u] + p` is less than `temp_prices[v]`, update `temp_prices[v]` to `prices[u] + p`.
Step 6: After iterating through all flights, update `prices` with the values from `temp_prices`. This completes one iteration, representing paths with at most `i` stops.
Step 7: After the loop finishes (after `k+1` iterations), check if `prices[dst]` is still infinity. If it is, it means there is no path from `src` to `dst` within `k` stops, so return -1.
Step 8: Otherwise, return `prices[dst]`, which represents the cheapest price to reach `dst` from `src` within `k` stops.

Key Insights

Insight 1: The core idea is to relax the edges of the graph representing the flight network `k+1` times. This relaxation process involves iterating through all flights and updating the minimum cost to reach the destination of each flight, considering the number of stops allowed.
Insight 2: The use of dynamic programming with a temporary array to store the updated prices after each iteration is essential. This prevents updating a price based on a price that has already been updated in the same iteration, which would violate the stop constraint.
Insight 3: Handling the 'no path' scenario by returning -1 when the minimum price to reach the destination remains infinity (or MAX_VALUE) after all iterations is crucial.

Complexity Analysis

Time Complexity: O(K * E)

Space Complexity: O(N)

Topics

This problem involves: Dynamic Programming, Depth-First Search, Breadth-First Search, Graph, Heap (Priority Queue), Shortest Path.

Companies

Asked at: Airbnb, Cloudera, Coupang, DE Shaw, MakeMyTrip, Snap, Snowflake, Stripe.

Cheapest Flights Within K Stops | LeetCode 787

Medium

Dynamic Programming Depth-First Search Breadth-First Search Graph Heap (Priority Queue)Shortest Path

Time: O(K * E)Space: O(N)

Solution Code

Available in:

class Solution {
    public int findCheapestPrice(int n, int[][] flights, int src, int dst, int k) {
        int[] prices = new int[n];
        for (int i = 0; i < n; i++) {
            prices[i] = Integer.MAX_VALUE;
        }
        prices[src] = 0;

        for (int i = 0; i < k + 1; i++) {
            int[] temp_prices = prices.clone();
            for (int[] flight : flights) {
                int u = flight[0];
                int v = flight[1];
                int p = flight[2];
                if (prices[u] != Integer.MAX_VALUE) {
                    if (prices[u] + p < temp_prices[v]) {
                        temp_prices[v] = prices[u] + p;
                    }
                }
            }
            prices = temp_prices;
        }

        if (prices[dst] == Integer.MAX_VALUE) {
            return -1;
        }

        return prices[dst];
    }
}

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Problem Info

Description

Problem 787: Cheapest Flights Within K Stops - Detailed explanation and solution approaches.

Topics

Dynamic Programming Depth-First Search Breadth-First Search Graph Heap (Priority Queue)Shortest Path

Companies

AirbnbClouderaCoupangDE ShawMakeMyTripSnapSnowflakeStripe

Stats

Difficulty:Medium

Languages:4

Companies:8

Problem #:787

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize an array `prices` of size `n` with infinity (or MAX_VALUE) for all cities except the source city, which is initialized to 0.
2
Step 2: Iterate `k+1` times (representing the maximum number of stops allowed).
3
Step 3: In each iteration, create a temporary copy `temp_prices` of the `prices` array. This is critical to avoid incorrect updates during the iteration.
4
Step 4: Iterate through the `flights` array. For each flight `[u, v, p]`, check if `prices[u]` is not infinity. If it's not, check if `prices[u] + p` (the cost to reach city `u` plus the cost of the flight from `u` to `v`) is less than `temp_prices[v]` (the current minimum cost to reach city `v`).
5
Step 5: If `prices[u] + p` is less than `temp_prices[v]`, update `temp_prices[v]` to `prices[u] + p`.
6
Step 6: After iterating through all flights, update `prices` with the values from `temp_prices`. This completes one iteration, representing paths with at most `i` stops.
7
Step 7: After the loop finishes (after `k+1` iterations), check if `prices[dst]` is still infinity. If it is, it means there is no path from `src` to `dst` within `k` stops, so return -1.
8
Step 8: Otherwise, return `prices[dst]`, which represents the cheapest price to reach `dst` from `src` within `k` stops.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(K * E)

💾

Space Complexity

O(N)

📝

Why?

The time complexity arises from iterating through all edges (flights) for each allowed number of stops. The space complexity is due to the auxillary arrays used for dynamic programming, which store the minimum prices to each city.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Breadth-First Search (BFS)

Use a queue to perform a BFS, keeping track of the number of stops taken. Store the cost to reach each city with a certain number of stops. In each step, explore the neighbors of the current city and update the cost if a cheaper path is found. This can be optimized using a priority queue to explore cheaper paths first, similar to Dijkstra's algorithm.

Time: O(V + E), where V is vertices (cities) and E is edges (flights), in the best case, if we implement using priority queue. O(V*E) in worst case if just using queue.Space: O(V + E), where V is vertices (cities) and E is edges (flights)

Trade-offs: BFS with priority queue can be faster if the graph is sparse and k is large. However, it can be more complex to implement. The dynamic programming approach is simpler to understand and implement.

Dijkstra's Algorithm with stops

A modification of Dijkstra's algorithm can be used. However, standard Dijkstra's does not allow the number of stops to be constrained. A modified version can be used that keeps track of the cost and number of stops to each node, and only enqueues nodes if they result in a cheaper cost with fewer stops than previously seen.

Time: O(E log V), where V is vertices (cities) and E is edges (flights). The log V comes from the use of the priority queue.Space: O(V + E), where V is vertices (cities) and E is edges (flights)

Trade-offs: Potentially faster than the main solution but only if implemented properly. However, it tends to be more complex to implement and debug.

✨

Key Insights

1Insight 1: The core idea is to relax the edges of the graph representing the flight network `k+1` times. This relaxation process involves iterating through all flights and updating the minimum cost to reach the destination of each flight, considering the number of stops allowed.
2Insight 2: The use of dynamic programming with a temporary array to store the updated prices after each iteration is essential. This prevents updating a price based on a price that has already been updated in the same iteration, which would violate the stop constraint.
3Insight 3: Handling the 'no path' scenario by returning -1 when the minimum price to reach the destination remains infinity (or MAX_VALUE) after all iterations is crucial.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: When there is no path from `src` to `dst` within `k` stops. The solution correctly returns -1 in this case because `prices[dst]` remains infinity after the iterations.
Edge case 2: When `k` is 0, meaning only direct flights are allowed. The solution correctly considers only direct flights from `src` during the first iteration and finds the minimum cost.
Edge case 3: When the graph contains negative cycles. While the problem statement doesn't explicitly exclude negative cycles, if present, Bellman-Ford (the underlying principle) would continue to relax edges even after k+1 iterations. This solution will not detect negative cycles and may return an incorrect result or overflow issues (handled by using long long in C when adding potentially very large numbers).
Edge case 4: Large values for price might lead to integer overflow when adding. C solution addresses this issue by using `long long` for computation.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a temporary array to store the updated prices in each iteration. This leads to incorrect updates because the prices of cities are updated based on prices that have already been updated in the same iteration, violating the stop constraint.

✗Common mistake 2: Not initializing the `prices` array correctly with infinity for all cities except the source city.

✗Common mistake 3: Not handling the case where there is no path from `src` to `dst` within `k` stops. Forgetting to return -1 in this scenario.

✗Common mistake 4: Integer overflow. The sum of prices along a path might exceed the maximum integer value. C addresses this by using `long long` for intermediate calculations.

✗Common mistake 5: Misunderstanding the number of iterations needed. The loop should run `k+1` times because k represents the maximum number of *stops*, which means the path can have at most `k+1` segments / edges.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Flight booking systems that need to find the cheapest flights between cities with a maximum number of layovers (stops).

🏦Real-world scenario 2: Network routing protocols that need to find the shortest path between two nodes in a network with a limited number of hops.

🎮Real-world scenario 3: Supply chain optimization, where finding the most cost-effective route for transporting goods between locations with constraints on the number of intermediate stops or transshipment points is crucial.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 7, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Cheapest Flights Within K Stops"?

For this medium-level Dynamic Programming problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?