Jump Game Problem in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the array is empty and return true.

DSA Java

if (nums.length [1] 0) return true;

Drag options to blanks, or click blank then click option'

A!=

B>

C<

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' or '<' instead of '==' causes wrong checks.

Using '!=' will return true for non-empty arrays.

We check if the length of the array is exactly 0 to return true immediately.

2fill in blank

medium

Complete the code to update the farthest reachable index.

DSA Java

farthest = Math.[1](farthest, i + nums[i]);

Drag options to blanks, or click blank then click option'

Amax

Bmin

Cabs

Dround

Attempts:

3 left

💡 Hint

Common Mistakes

Using Math.min will incorrectly reduce the farthest index.

Using Math.abs or Math.round is unrelated to this logic.

We use Math.max to keep track of the farthest index we can jump to.

3fill in blank

hard

Fix the error in the loop condition to avoid index out of bounds.

DSA Java

for (int i = 0; i [1] nums.length; i++) {

Drag options to blanks, or click blank then click option'

A<=

B<

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes the loop to access an invalid index.

Using '>' or '>=' will not run the loop correctly.

The loop should run while i is less than nums.length to avoid going out of bounds.

4fill in blank

hard

Fill both blanks to check if current index is reachable and update farthest jump.

DSA Java

if (i [1] farthest) {
    farthest = Math.[2](farthest, i + nums[i]);
}

Drag options to blanks, or click blank then click option'

A<=

B<

Cmax

Dmin

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '<=' may skip valid indices.

Using Math.min will incorrectly reduce farthest.

We check if i is less than or equal to farthest reachable index, then update farthest using Math.max.

5fill in blank

hard

Fill all three blanks to return true if last index is reachable, else false.

DSA Java

return farthest [1] nums.length [2] 1;

// or

return farthest [3] nums.length - 1;

Drag options to blanks, or click blank then click option'

A>=

B<=

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' will give wrong result.

Using '+' instead of '-' for last index is incorrect.

We check if farthest is greater than or equal to last index (nums.length - 1). The '+' in second line is part of expression 'nums.length + 1' which is incorrect, so correct is to use '-' in actual code. Here, the blanks show the correct operators to form the condition.

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