Jump Game Problem in DSA Java - Time & Space Complexity

Time Complexity: Jump Game Problem

O(n)

Understanding Time Complexity

We want to know how the time needed to solve the Jump Game problem changes as the input grows.

Specifically, how does the number of steps to check jumps grow with the size of the input array?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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;
}
    

This code checks if you can jump from the start to the end of the array by updating the farthest reachable index.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop that goes through each element of the array once.
• How many times: The loop runs once for each element, so n times if the array length is n.

How Execution Grows With Input

As the array size grows, the number of steps grows roughly the same way.

Input Size (n)	Approx. Operations
10	About 10 checks
100	About 100 checks
1000	About 1000 checks

Pattern observation: The work grows directly with the input size, so doubling the input roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to decide if you can jump to the end grows in a straight line with the size of the input array.

Common Mistake

[X] Wrong: "Because we check jumps inside the loop, the time is more than linear, maybe quadratic."

[OK] Correct: The code only updates the farthest reachable index once per element and never rechecks previous elements, so it does not multiply work inside the loop.

Interview Connect

Understanding this linear time solution shows you can analyze and optimize problems involving array traversal and reachability, a common skill in interviews.

Self-Check

"What if we used a nested loop to check all possible jumps from each position? How would the time complexity change?"

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