What is the time complexity of Minimum Sideway Jumps?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum Sideway Jumps?

This problem uses Array, Dynamic Programming, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Sideway Jumps?

Insight 1: Dynamic programming is suitable because the minimum number of jumps to reach a lane at a given point depends on the minimum number of jumps to reach other lanes at the previous point. Insight 2: We can optimize space by only storing the minimum jumps for the current point, as the values from previous points are no longer needed after processing.

Which companies ask LeetCode 1824. Minimum Sideway Jumps in interviews?

This problem is frequently asked at Pony.ai.

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Minimum Sideway Jumps - LeetCode 1824 Solution

Minimum Sideway Jumps - Complete Solution Guide

Minimum Sideway Jumps is LeetCode problem 1824, 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 is a 3 lane road of length n that consists of n + 1 points labeled from 0 to n . A frog starts at point 0 in the second lane and wants to jump to point n . However, there could be obstacles along the way. You are given an array obstacles of length n + 1 where each obstacles[i] ( ranging from 0 to 3 ) describes an obstacle on the lane obstacles[i] at point i . If obstacles[i] == 0 , there are no obstacles at point i . There will be at most one obstacle in the 3 lanes at each point. For exam

Detailed Explanation

The problem describes a frog trying to reach the end of a 3-lane road with obstacles. The road is divided into points numbered from 0 to n. The frog starts at lane 2 (index 1) at point 0 and wants to reach any lane at point n. The input `obstacles` array specifies which lane, if any, has an obstacle at each point. The frog can move forward one point on the same lane if there is no obstacle. It can also make side jumps to any other lane at the same point, regardless of adjacency, as long as that lane isn't blocked by an obstacle at that point. The task is to find the minimum number of side jumps needed for the frog to reach point n.

Solution Approach

The solution uses dynamic programming to find the minimum number of side jumps. `dp[i]` represents the minimum side jumps needed to reach lane `i+1` at the current point. We initialize `dp` based on the frog's starting position. We then iterate through each point, updating the `dp` values based on whether there are obstacles and considering all possible side jumps. To avoid overwriting and incorrectly using partial information, we store copies of previous values, then update the array after processing the current point.

Step-by-Step Algorithm

Step 1: Initialize a `dp` array of size 3. `dp[0] = 1`, `dp[1] = 0`, and `dp[2] = 1`, reflecting the initial number of jumps needed to be on lane 1, 2 and 3 respectively at position 0.
Step 2: Iterate through the `obstacles` array from index 1 (representing point 1) to the end.
Step 3: In each iteration, check if there's an obstacle on any of the lanes. If `obstacles[i] == lane`, then `dp[lane - 1] = infinity` (or Integer.MAX_VALUE in Java/C++) because that lane is blocked and not reachable via a forward move.
Step 4: Create copies of the current `dp` values. In Python the values are c0, c1 and c2.
Step 5: Recalculate the `dp` values for each lane, taking into account possible side jumps. `dp[i] = min(dp[i], dp[j] + 1, dp[k] + 1)` where i, j, and k are the lane indices (0, 1, 2) and j and k represent the other lanes. If a lane `i` has an obstacle at the current point, its dp value is not updated in this step, and instead uses the infinite value previously set.
Step 6: Repeat steps 3-5 for all points.
Step 7: After iterating through all points, return the minimum value in the `dp` array.

Key Insights

Insight 1: Dynamic programming is suitable because the minimum number of jumps to reach a lane at a given point depends on the minimum number of jumps to reach other lanes at the previous point.
Insight 2: We can optimize space by only storing the minimum jumps for the current point, as the values from previous points are no longer needed after processing.
Insight 3: When a lane has an obstacle at a certain point, that lane becomes unreachable through a forward move, so its cost should be set to infinity.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Greedy.

Companies

Asked at: Pony.ai.

Minimum Sideway Jumps | LeetCode 1824

Medium

Array Dynamic Programming Greedy

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minSideJumps(int[] obstacles) {
        // dp[j] will store the minimum side jumps to reach the current point `i` on lane `j+1`.
        // Initialized for point 0. Frog starts at lane 2 (0 jumps).
        // To be on lane 1 or 3 requires 1 side jump.
        int[] dp = {1, 0, 1};

        // Iterate through points from 1 to n.
        // obstacles[0] is for point 0, which has no obstacles and is handled by initialization.
        for (int obs : obstacles) {
            // If there's an obstacle on a lane at the current point,
            // that lane is unreachable by a forward move.
            if (obs > 0) {
                dp[obs - 1] = Integer.MAX_VALUE;
            }

            // Store the costs before considering side jumps at the current point
            // to avoid using partially updated values in the same step.
            int c0 = dp[0], c1 = dp[1], c2 = dp[2];

            // Recalculate costs for each lane including side jumps at the current point.
            // A lane is only a potential destination if it's not blocked.
            if (obs != 1) {
                dp[0] = Math.min(c0, Math.min(c1 + 1, c2 + 1));
            }
            if (obs != 2) {
                dp[1] = Math.min(c1, Math.min(c0 + 1, c2 + 1));
            }
            if (obs != 3) {
                dp[2] = Math.min(c2, Math.min(c0 + 1, c1 + 1));
            }
        }

        return Math.min(dp[0], Math.min(dp[1], dp[2]));
    }
}

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Problem Info

Description

Problem 1824: Minimum Sideway Jumps - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Greedy

Companies

Pony.ai

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1824

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

7 steps

1
Step 1: Initialize a `dp` array of size 3. `dp[0] = 1`, `dp[1] = 0`, and `dp[2] = 1`, reflecting the initial number of jumps needed to be on lane 1, 2 and 3 respectively at position 0.
2
Step 2: Iterate through the `obstacles` array from index 1 (representing point 1) to the end.
3
Step 3: In each iteration, check if there's an obstacle on any of the lanes. If `obstacles[i] == lane`, then `dp[lane - 1] = infinity` (or Integer.MAX_VALUE in Java/C++) because that lane is blocked and not reachable via a forward move.
4
Step 4: Create copies of the current `dp` values. In Python the values are c0, c1 and c2.
5
Step 5: Recalculate the `dp` values for each lane, taking into account possible side jumps. `dp[i] = min(dp[i], dp[j] + 1, dp[k] + 1)` where i, j, and k are the lane indices (0, 1, 2) and j and k represent the other lanes. If a lane `i` has an obstacle at the current point, its dp value is not updated in this step, and instead uses the infinite value previously set.
6
Step 6: Repeat steps 3-5 for all points.
7
Step 7: After iterating through all points, return the minimum value in the `dp` array.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The outer loop iterates through the obstacles array which has 'n' elements. Inside the loop, only constant time operations are performed like array access, min calculations, etc. Hence, the time complexity is O(n). Only a fixed size array of 3 integers is used, making the space complexity O(1).

📖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

Graph Traversal (BFS/DFS)

Represent the problem as a graph where each node is a tuple (point, lane). Edges connect adjacent points on the same lane (if no obstacle) and lanes at the same point (side jumps). Use BFS or DFS to find the shortest path from the starting node (0, 2) to any node (n, lane).

Time: O(n), assuming a limited number of edges from each node, and since obstacles limit the total edges.Space: O(n) in the worst case, to store the graph (adjacency list) and the queue/stack for BFS/DFS.

Trade-offs: Graph traversal is more intuitive but potentially uses more memory than dynamic programming for this problem, given the constraints. The dynamic programming solution is simpler to implement and more efficient regarding memory usage.

✨

Key Insights

1Insight 1: Dynamic programming is suitable because the minimum number of jumps to reach a lane at a given point depends on the minimum number of jumps to reach other lanes at the previous point.
2Insight 2: We can optimize space by only storing the minimum jumps for the current point, as the values from previous points are no longer needed after processing.
3Insight 3: When a lane has an obstacle at a certain point, that lane becomes unreachable through a forward move, so its cost should be set to infinity.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: No obstacles at all. The algorithm correctly returns 0 if the frog can reach the end on lane 2 without any jumps.
Edge case 2: An obstacle blocks the starting lane (lane 2) at some point. The algorithm correctly handles this by considering side jumps to other lanes and keeping track of the minimum jumps.
Edge case 3: An obstacle appears right at the starting point. The problem statement guarantees obstacles[0] == 0, and obstacles[n] == 0 so this is not possible. The code explicitly starts iteration at obstacles[1].
Edge case 4: All lanes are blocked at some point. The `dp` values will become infinity. The algorithm will find the minimum of all lanes to reach the destination.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the obstacles correctly. Failing to set the cost of a blocked lane to infinity leads to incorrect results.

✗Common mistake 2: Not considering all possible side jumps when updating the `dp` array. It's important to compare the current cost of a lane with the cost of reaching it from any other lane plus one jump.

✗Common mistake 3: Using outdated values in `dp` calculation. Creating temporary copies before updating the array is crucial for correctness.

✅

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: Pathfinding in a video game where a character needs to navigate through a level with obstacles and can switch between different paths/lanes to avoid them.

🏦Real-world scenario 2: Route planning for autonomous vehicles in a multi-lane road system with dynamically changing lane closures or obstacles. The algorithm can help determine the optimal lane changes to reach the destination with minimal switches.

🎮Real-world scenario 3: Network routing where packets need to be routed through different paths, with some paths having congestion or failures. The algorithm can help determine the optimal path with the fewest number of switches.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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: June 25, 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

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

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Sideway Jumps"?

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

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