Jump Game Problem in DSA Python - Time & Space Complexity

Time Complexity: Jump Game Problem

O(n)

Understanding Time Complexity

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

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def can_jump(nums):
    max_reach = 0
    for i in range(len(nums)):
        if i > max_reach:
            return False
        max_reach = max(max_reach, i + nums[i])
    return True

This code checks if you can jump from the start to the end of the array, where each number tells the max jump length from that position.

Identify Repeating Operations

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

How Execution Grows With Input

As the array gets longer, the code checks each position once, so the work grows directly with the input size.

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

Pattern observation: The number of operations grows in a straight line with the input size.

Final Time Complexity

Time Complexity: O(n)

This means the time to decide if you can jump to the end grows directly with the number of positions you have to check.

Common Mistake

[X] Wrong: "Because we check jumps from each position, the time is O(n²) because of nested jumps."

[OK] Correct: The code only loops once through the array and updates the farthest reachable position without nested loops, so it does not multiply checks.

Interview Connect

Understanding this linear time solution shows you can analyze and optimize problems that look like they might need more work, a skill valued in interviews and real projects.

Self-Check

"What if we changed the code to try all possible jumps from each position instead of just tracking the farthest reach? How would the time complexity change?"

Practice

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

easy

A. The maximum jump length possible from the first index.

B. If you can reach the last index starting from the first index by jumping according to the values in nums.

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 numbers in the array as maximum jump lengths.

2. Step 2: Identify the goal

   The goal is to check if the last index is reachable, not how many jumps it takes or the sum of jumps.

3. Final Answer:

   If you can reach the last index starting from the first index by jumping according to the values in nums. -> Option B

4. Quick Check:

   Jump Game = Reach last index [OK]

Hint: Focus on reachability, not jump count or sums [OK]

Common Mistakes:

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

2. Which of the following Python code snippets correctly checks if you can jump to the last index in the Jump Game problem?

easy

A. max_reach = 0
for i in range(len(nums)):
if i > max_reach:
return False
max_reach = max(max_reach, i + nums[i])
return True

B. for i in range(len(nums)): if i < max_reach: return False max_reach = max(max_reach, i + nums[i]) return True

C. for i in range(len(nums)): max_reach = min(max_reach, i + nums[i]) return max_reach >= len(nums)-1

D. for i in range(len(nums)): if i > max_reach: return True max_reach = max(max_reach, i + nums[i]) return False

Solution

1. Step 1: Understand the logic of max_reach

   We track the farthest index reachable so far. If current index is beyond max_reach, we cannot proceed.

2. Step 2: Analyze the code snippets

   for i in range(len(nums)): if i > max_reach: return False max_reach = max(max_reach, i + nums[i]) return True correctly returns False if current index is beyond max_reach, updates max_reach, and returns True if loop finishes.

3. Final Answer:

   Option A code snippet correctly implements the jump check logic. -> Option A

4. Quick Check:

   Check i > max_reach to fail early [OK]

Hint: Check if current index exceeds max reachable index [OK]

Common Mistakes:

• Using < instead of > to compare index and max_reach
• Returning True or False too early inside loop
• Using min instead of max for max_reach

3. What is the output of the following code for nums = [2,3,1,1,4]?

def can_jump(nums):
    max_reach = 0
    for i in range(len(nums)):
        if i > max_reach:
            return False
        max_reach = max(max_reach, i + nums[i])
    return True

print(can_jump(nums))

medium

A. IndexError

B. False

C. True

D. None

Solution

1. Step 1: Trace the loop and max_reach updates

   Start max_reach=0. i=0: 0 <= 0, max_reach = max(0,0+2)=2. i=1: 1 <= 2, max_reach = max(2,1+3)=4. i=2: 2 <= 4, max_reach = max(4,2+1)=4. i=3: 3 <= 4, max_reach = max(4,3+1)=4. i=4: 4 <= 4, max_reach = max(4,4+4)=8.

2. Step 2: Confirm loop completes and returns True

   Since loop never returns False, function returns True at end.

3. Final Answer:

   True -> Option C

4. Quick Check:

   max_reach covers last index, so True [OK]

Hint: Update max_reach and check if current index is reachable [OK]

Common Mistakes:

• Stopping early without checking all indices
• Confusing max and min when updating max_reach
• Assuming False if max_reach equals current index

4. The following code is intended to solve the Jump Game problem but has a bug. What is the bug?

def can_jump(nums):
    max_reach = 0
    for i in range(len(nums)):
        if i >= max_reach:
            return False
        max_reach = max(max_reach, i + nums[i])
    return True

medium

A. The loop should start from i = 1, not 0.

B. The max_reach update should use min instead of max.

C. The function should return False at the end, not True.

D. The condition should be 'if i > max_reach' instead of 'if i >= max_reach'.

Solution

1. Step 1: Analyze the condition 'if i >= max_reach'

   This condition incorrectly returns False even when i equals max_reach, which is a reachable index.

2. Step 2: Correct the condition to 'if i > max_reach'

   Changing to '>' allows the current index to be equal to max_reach, meaning it is reachable and should not return False prematurely.

3. Final Answer:

   The condition should be 'if i > max_reach' instead of 'if i >= max_reach'. -> Option D

4. Quick Check:

   Use strict greater than to detect unreachable index [OK]

Hint: Use '>' not '>=' to check if index is beyond reach [OK]

Common Mistakes:

• Using '>=' causes false negatives
• Updating max_reach incorrectly
• Returning False too early

5. Given the array nums = [3,2,1,0,4], why does the standard greedy Jump Game solution return False?

hard

A. Because the zero at index 3 blocks any further jumps, making the last index unreachable.

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

C. Because the array length is less than 5, so the problem is invalid.

D. Because the last element is greater than the sum of previous elements.

Solution

1. Step 1: Trace jumps and max_reach

   Start max_reach=0. i=0: max_reach = max(0,0+3)=3. i=1: 1 <= 3, max_reach = max(3,1+2)=3. i=2: 2 <= 3, max_reach = max(3,2+1)=3. i=3: 3 <= 3, max_reach = max(3,3+0)=3. i=4: 4 > 3, so return False.

2. Step 2: Understand why last index is unreachable

   The zero at index 3 prevents extending max_reach beyond 3, so index 4 cannot be reached.

3. Final Answer:

   Because the zero at index 3 blocks any further jumps, making the last index unreachable. -> Option A

4. Quick Check:

   Zero blocks jump, last index unreachable [OK]

Hint: Zeros can block progress if max_reach can't pass them [OK]

Common Mistakes:

• Ignoring zeros that block jumps
• Assuming large first element guarantees success
• Misunderstanding array length relevance