Jump Game Problem in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Jump Game Problem?

To determine if you can reach the last index of an array starting from the first index, where each element represents the maximum jump length from that position.

beginner

In the Jump Game Problem, what does each element in the input array represent?

Each element represents the maximum number of steps you can jump forward from that position.

intermediate

What is a greedy approach to solve the Jump Game Problem?

Keep track of the farthest reachable index while iterating through the array. If at any point the current index is beyond the farthest reachable, return false. If you can reach or pass the last index, return true.

intermediate

Why does the greedy approach work for the Jump Game Problem?

Because always jumping to the farthest reachable position ensures you don't miss any possible path to the end, making it efficient and correct.

beginner

What is the time complexity of the greedy solution for the Jump Game Problem?

O(n), where n is the length of the input array, since it requires a single pass through the array.

In the Jump Game Problem, what does a zero in the array mean?

A You cannot jump forward from that position

B You must jump exactly one step

C You can jump any number of steps

D You can jump backward

A zero means you cannot move forward from that position, which might block progress if you land there.

What should you do if your current index is greater than the farthest reachable index in the greedy approach?

A Continue iterating

B Return true

C Return false

D Reset the farthest reachable index

If current index is beyond farthest reachable, you cannot move forward, so return false.

What is the initial value of the farthest reachable index in the greedy solution?

A 1

B -1

C Length of array

D 0

We start at index 0, so initially the farthest reachable index is 0.

If the input array is [2,3,1,1,4], can you reach the last index?

A Yes

B No

C Only if you jump backward

D Only if you skip some elements

You can jump from index 0 to 1 (max 2 steps), then from 1 to 4, reaching the last index.

What is the space complexity of the greedy solution for the Jump Game Problem?

A O(n)

B O(1)

C O(log n)

D O(n^2)

The greedy solution uses constant extra space, so space complexity is O(1).

Explain how the greedy algorithm solves the Jump Game Problem step-by-step.

Think about how you move forward and keep track of the best place you can jump to.

Describe a scenario where you cannot reach the last index in the Jump Game Problem and why.

Consider what happens if you land on a zero and cannot jump further.

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