What is the time complexity of Jump Game III?

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

What data structures are used to solve Jump Game III?

This problem uses Array, Depth-First Search, Breadth-First Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Jump Game III?

Insight 1: The problem can be modeled as a graph traversal problem where array indices are nodes and jumps are edges. Insight 2: We can use Breadth-First Search (BFS) or Depth-First Search (DFS) to explore the possible jump paths from the starting index.

Which companies ask LeetCode 1306. Jump Game III in interviews?

This problem is frequently asked at Pinterest, Snap, Tanium.

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Jump Game III - LeetCode 1306 Solution

Jump Game III - Complete Solution Guide

Jump Game III is LeetCode problem 1306, 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

Given an array of non-negative integers arr , you are initially positioned at start index of the array. When you are at index i , you can jump to i + arr[i] or i - arr[i] , check if you can reach any index with value 0. Notice that you can not jump outside of the array at any time. Example 1: Input: arr = [4,2,3,0,3,1,2], start = 5 Output: true Explanation: All possible ways to reach at index 3 with value 0 are: index 5 -> index 4 -> index 1 -> index 3 index 5 -> index 6 -> index 4 -> index 1 ->

Detailed Explanation

The problem, Jump Game III, presents an array of non-negative integers, `arr`, and a starting index, `start`. The goal is to determine if it's possible to reach any index within the array that holds the value 0, by making jumps. From any index `i`, you can jump to either `i + arr[i]` or `i - arr[i]`, but you cannot jump outside the bounds of the array. The problem requires us to find *any* path leading to a zero value, not necessarily the shortest or most efficient one. The constraints include the array size, element values, and the starting index, all within reasonable bounds.

Solution Approach

The provided code uses a Breadth-First Search (BFS) algorithm to explore the array and find a path to an index with value 0. It initializes a queue with the starting index and a set to keep track of visited indices. The algorithm iteratively dequeues an index, checks if its value is 0, and if not, adds its valid, unvisited neighbors (the indices reachable by jumping forward and backward) to the queue and marks them as visited. The process continues until a 0 is found or the queue is empty, indicating that no path to 0 exists from the starting point.

Step-by-Step Algorithm

Step 1: Check if the value at the starting index is 0. If it is, return `true` immediately.
Step 2: Initialize a queue with the starting index and a set to store visited indices, adding the starting index to the visited set.
Step 3: While the queue is not empty, dequeue an index `idx` from the queue.
Step 4: Check if `arr[idx]` is equal to 0. If it is, return `true`.
Step 5: Calculate the next possible index by jumping forward: `next_idx_forward = idx + arr[idx]`.
Step 6: If `next_idx_forward` is within the array bounds and has not been visited, add it to the queue and mark it as visited.
Step 7: Calculate the next possible index by jumping backward: `next_idx_backward = idx - arr[idx]`.
Step 8: If `next_idx_backward` is within the array bounds and has not been visited, add it to the queue and mark it as visited.
Step 9: Repeat steps 3-8 until the queue is empty.
Step 10: If the loop finishes without finding an index with value 0, return `false`.

Key Insights

Insight 1: The problem can be modeled as a graph traversal problem where array indices are nodes and jumps are edges.
Insight 2: We can use Breadth-First Search (BFS) or Depth-First Search (DFS) to explore the possible jump paths from the starting index.
Insight 3: To prevent infinite loops, we need to keep track of visited indices to avoid revisiting them.
Insight 4: Optimizing the initial check for arr[start] == 0 improves code readability and efficiency.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Depth-First Search, Breadth-First Search.

Companies

Asked at: Pinterest, Snap, Tanium.

Jump Game III | LeetCode 1306

Medium

Array Depth-First Search Breadth-First Search

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

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.HashSet;
import java.util.Queue;
import java.util.Set;

class Solution {
    public boolean canReach(int[] arr, int start) {
        int n = arr.length;
        if (0 <= start && start < n && arr[start] == 0) {
            return true;
        }

        Queue<Integer> q = new ArrayDeque<>();
        q.offer(start);
        Set<Integer> visited = new HashSet<>();
        visited.add(start);

        while (!q.isEmpty()) {
            int idx = q.poll();

            if (arr[idx] == 0) {
                return true;
            }

            // Jump forward
            int next_idx_forward = idx + arr[idx];
            if (next_idx_forward < n && !visited.contains(next_idx_forward)) {
                visited.add(next_idx_forward);
                q.offer(next_idx_forward);
            }

            // Jump backward
            int next_idx_backward = idx - arr[idx];
            if (next_idx_backward >= 0 && !visited.contains(next_idx_backward)) {
                visited.add(next_idx_backward);
                q.offer(next_idx_backward);
            }

        }

        return false;
    }
}

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Problem Info

Description

Problem 1306: Jump Game III - Detailed explanation and solution approaches.

Topics

Array Depth-First Search Breadth-First Search

Companies

PinterestSnapTanium

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:1306

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

10 steps

1
Step 1: Check if the value at the starting index is 0. If it is, return `true` immediately.
2
Step 2: Initialize a queue with the starting index and a set to store visited indices, adding the starting index to the visited set.
3
Step 3: While the queue is not empty, dequeue an index `idx` from the queue.
4
Step 4: Check if `arr[idx]` is equal to 0. If it is, return `true`.
5
Step 5: Calculate the next possible index by jumping forward: `next_idx_forward = idx + arr[idx]`.
6
Step 6: If `next_idx_forward` is within the array bounds and has not been visited, add it to the queue and mark it as visited.
7
Step 7: Calculate the next possible index by jumping backward: `next_idx_backward = idx - arr[idx]`.
8
Step 8: If `next_idx_backward` is within the array bounds and has not been visited, add it to the queue and mark it as visited.
9
Step 9: Repeat steps 3-8 until the queue is empty.
10
Step 10: If the loop finishes without finding an index with value 0, return `false`.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because in the worst-case scenario, we might enqueue and dequeue each element of the array once. The space complexity is O(n) because the visited set could potentially store all the indices in the array, especially if all elements are reachable.

📖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

Depth-First Search (DFS)

Instead of using a queue, we can use recursion to explore the possible jump paths. We would need a visited set to avoid infinite recursion.

Time: O(n)Space: O(n) in the worst case due to the recursion depth and the visited set.

Trade-offs: DFS can lead to stack overflow errors for very large arrays if the recursion depth becomes too large. BFS is generally preferred in this case because it explores the graph level by level and avoids deep recursion.

✨

Key Insights

1Insight 1: The problem can be modeled as a graph traversal problem where array indices are nodes and jumps are edges.
2Insight 2: We can use Breadth-First Search (BFS) or Depth-First Search (DFS) to explore the possible jump paths from the starting index.
3Insight 3: To prevent infinite loops, we need to keep track of visited indices to avoid revisiting them.
4Insight 4: Optimizing the initial check for arr[start] == 0 improves code readability and efficiency.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: The starting index already contains a 0. The code handles this by immediately returning `true` in the initial check, avoiding unnecessary traversal.
Edge case 2: The array contains only one element, and that element is not 0. The algorithm will check the starting index, and the queue will be empty, causing it to correctly return `false`.
Edge case 3: The array contains loops, meaning the jumps might lead back to previously visited indices. The `visited` set prevents infinite loops by ensuring each index is visited only once.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a `visited` set or equivalent mechanism, leading to infinite loops and incorrect results, especially in arrays with cyclic jump patterns.

✗Common mistake 2: Not handling the array bounds correctly, which can lead to `IndexOutOfBoundsException` or similar errors. The solutions ensure that `next_idx_forward` is less than `n` and `next_idx_backward` is greater than or equal to `0` before processing them.

✗Common mistake 3: Confusing the problem with finding the *shortest* path to 0. The problem only asks if *any* path exists, so simpler graph traversal algorithms are sufficient.

✅

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: Network routing, where each node represents a router and `arr[i]` represents the distance to a neighboring router. The problem becomes finding if a path exists to a router with a specific configuration (e.g., a router connected to the internet).

🏦Real-world scenario 2: Game development, where the array represents possible moves in a game. The problem is determining if it's possible to reach a winning state (represented by a 0 value) from a given starting position.

🎮Real-world scenario 3: Solving mazes, where the array represents the possible paths one can take. The objective is to find if there is a path leading to the exit, denoted by 0.

💡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

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

⚠️ 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: July 17, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

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

@InterviewProJul 6

❓

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.

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 "Jump Game III"?

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 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?