Jump Game Problem in DSA Java - Build from Scratch

Jump Game Problem

📖 Scenario: Imagine you are playing a video game where you need to jump across platforms to reach the end. Each platform lets you jump forward a certain number of steps. You want to check if you can reach the last platform starting from the first one.

🎯 Goal: Build a Java program that checks if you can jump from the first platform to the last platform using the jump lengths given in an array.

📋 What You'll Learn

Create an integer array called jumps with the exact values {2, 3, 1, 1, 4}.

Create an integer variable called maxReach and set it to 0.

Use a for loop with variable i to iterate over the jumps array and update maxReach to track the farthest platform reachable.

Print "Can reach the last platform: true" if the last platform is reachable, otherwise print "Can reach the last platform: false".

💡 Why This Matters

🌍 Real World

This problem models scenarios like navigating through levels in a game or planning steps in a path where each step has a limited range.

💼 Career

Understanding this problem helps in software development roles that require problem-solving with arrays and loops, such as game development, pathfinding algorithms, and optimization tasks.

1

Create the jumps array

Create an integer array called jumps with these exact values: {2, 3, 1, 1, 4}.

DSA Java

// Your code here

Hint

Use int[] jumps = {2, 3, 1, 1, 4}; to create the array.

2

Create maxReach variable

Create an integer variable called maxReach and set it to 0 inside the main method after the jumps array.

DSA Java

public class JumpGame {
    public static void main(String[] args) {
        int[] jumps = {2, 3, 1, 1, 4};
        // Your code here
    }
}

Hint

Declare int maxReach = 0; to track the farthest reachable platform.

3

Implement the jump logic

Use a for loop with variable i to iterate over the jumps array. Inside the loop, check if i is greater than maxReach. If yes, break the loop. Otherwise, update maxReach to the maximum of maxReach and i + jumps[i].

DSA Java

public class JumpGame {
    public static void main(String[] args) {
        int[] jumps = {2, 3, 1, 1, 4};
        int maxReach = 0;
        // Your code here
    }
}

Hint

Use for (int i = 0; i < jumps.length; i++) and update maxReach inside the loop.

4

Print if last platform is reachable

Print "Can reach the last platform: true" if maxReach is greater than or equal to jumps.length - 1. Otherwise, print "Can reach the last platform: false".

DSA Java

public class JumpGame {
    public static void main(String[] args) {
        int[] jumps = {2, 3, 1, 1, 4};
        int maxReach = 0;
        for (int i = 0; i  maxReach) {
                break;
            }
            maxReach = Math.max(maxReach, i + jumps[i]);
        }
        // Your code here
    }
}

Hint

Use System.out.println to print the result based on maxReach.

Practice

1. What does the Jump Game problem ask you to determine given an array of non-negative integers?

easy

A. If you can reach the last index starting from the first index by jumping according to array values

B. The maximum jump length you can make from the first index

C. The minimum number of jumps needed to reach the last index

D. The sum of all jump lengths in the array

Solution

1. Step 1: Understand the problem statement

   The Jump Game problem asks if you can start at the first index and reach the last index by jumping forward using the values at each index as maximum jump lengths.

2. Step 2: Identify the goal

   The goal is to check if reaching the last index is possible, not the number of jumps or sums.

3. Final Answer:

   If you can reach the last index starting from the first index by jumping according to array values -> Option A

4. Quick Check:

   Jump Game = Reach last index? [OK]

Hint: Jump Game checks reachability, not jump count or sums [OK]

Common Mistakes:

• Confusing reachability with minimum jumps
• Thinking problem asks for maximum jump length
• Assuming sum of jumps is required

2. Which of the following Java method signatures correctly represents a function to solve the Jump Game problem?

easy

A. public void canJump(int[] nums)

B. public int canJump(int[] nums)

C. public boolean canJump(int[] nums)

D. public String canJump(int[] nums)

Solution

1. Step 1: Determine the return type

   The Jump Game problem returns true or false depending on whether the last index is reachable, so the return type should be boolean.

2. Step 2: Check method signature correctness

   The method should accept an integer array and return a boolean, matching public boolean canJump(int[] nums).

3. Final Answer:

   public boolean canJump(int[] nums) -> Option C

4. Quick Check:

   Return type boolean for reachability [OK]

Hint: Jump Game method returns boolean for reachability [OK]

Common Mistakes:

• Using int return type instead of boolean
• Using void return type without returning result
• Returning String instead of boolean

3. What is the output of the following Java code snippet?

int[] nums = {2, 3, 1, 1, 4};
System.out.println(canJump(nums));

public static boolean canJump(int[] nums) {
    int maxReach = 0;
    for (int i = 0; i < nums.length; i++) {
        if (i > maxReach) return false;
        maxReach = Math.max(maxReach, i + nums[i]);
    }
    return true;
}

medium

A. Runtime exception

B. false

C. Compilation error

D. true

Solution

1. Step 1: Trace the loop and maxReach updates

   Start with maxReach=0. At i=0, maxReach=max(0,0+2)=2. i=1 ≤ 2, maxReach=max(2,1+3)=4. i=2 ≤ 4, maxReach=max(4,2+1)=4. i=3 ≤ 4, maxReach=max(4,3+1)=4. i=4 ≤ 4, maxReach=max(4,4+4)=8.

2. Step 2: Check if loop completes without returning false

   Since i never exceeds maxReach, loop completes and returns true.

3. Final Answer:

   true -> Option D

4. Quick Check:

   maxReach covers all indices, so true [OK]

Hint: Track maxReach; if index > maxReach, return false [OK]

Common Mistakes:

• Stopping early without updating maxReach
• Confusing maxReach with current index
• Ignoring the condition i > maxReach

4. Identify the bug in the following Java code for the Jump Game problem:

public boolean canJump(int[] nums) {
    int maxReach = 0;
    for (int i = 0; i <= nums.length; i++) {
        if (i > maxReach) return false;
        maxReach = Math.max(maxReach, i + nums[i]);
    }
    return true;
}

medium

A. The loop condition should be 'i < nums.length' instead of 'i <= nums.length' to avoid ArrayIndexOutOfBoundsException.

B. The maxReach update is incorrect; it should subtract 1 from i + nums[i].

C. The return type should be int, not boolean.

D. The method should return false at the end instead of true.

Solution

1. Step 1: Analyze the loop condition

   The loop runs from i=0 to i <= nums.length, which means it tries to access nums[nums.length], causing an ArrayIndexOutOfBoundsException.

2. Step 2: Correct the loop condition

   The loop should run while i < nums.length to stay within array bounds.

3. Final Answer:

   The loop condition should be 'i < nums.length' instead of 'i <= nums.length' to avoid ArrayIndexOutOfBoundsException. -> Option A

4. Quick Check:

   Array index must be less than length [OK]

Hint: Use i < nums.length to avoid array index errors [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Incorrect maxReach calculation
• Wrong return type for method

5. Given the array nums = {3, 2, 1, 0, 4}, why does the standard greedy Jump Game algorithm return false, and what does this imply about the array?

hard

A. Because the array length is too short to jump anywhere

B. Because index 3 has zero jump length blocking progress, making the last index unreachable

C. Because the first element is too large, causing an overflow error

D. Because the algorithm only works for arrays with all positive numbers

Solution

1. Step 1: Trace maxReach through the array

   Start maxReach=0. i=0: maxReach=max(0,0+3)=3. i=1 ≤ 3: maxReach=max(3,1+2)=3. i=2 ≤ 3: maxReach=max(3,2+1)=3. i=3 ≤ 3: maxReach=max(3,3+0)=3. i=4 > 3, so return false.

2. Step 2: Understand the implication

   Index 3 has zero jump length, blocking further progress, so the last index (4) is unreachable.

3. Final Answer:

   Because index 3 has zero jump length blocking progress, making the last index unreachable -> Option B

4. Quick Check:

   Zero jump blocks path, so false [OK]

Hint: Zero jump at blocking index means last index unreachable [OK]

Common Mistakes:

• Assuming large first element guarantees success
• Ignoring zero jump blocking
• Thinking algorithm fails on zeros always