Jump Game Problem in DSA Python - Execution Trace

Concept Flow - Jump Game Problem

Start at index 0

↓

Check max jump from current index

↓

Can we reach or pass last index?

Yes→Return True

No↓

Move to next reachable index

↓

Repeat until end or stuck

↓

If stuck before last index

→Return False

Start from the first position, check how far you can jump, move forward to the farthest reachable index, repeat until you reach or pass the last index or get stuck.

Execution Sample

DSA Python

nums = [2,3,1,1,4]
max_reach = 0
for i in range(len(nums)):
    if i > max_reach:
        print(False)
        break
    max_reach = max(max_reach, i + nums[i])
else:
    print(True)

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

Execution Table

Step	Operation	Current Index (i)	Max Reach	Condition i <= Max Reach?	Action	Visual State
1	Start loop	0	0	0 <= 0 (True)	Update max_reach = max(0, 0+2) = 2	[Index 0:2, Max Reach:2]
2	Next iteration	1	2	1 <= 2 (True)	Update max_reach = max(2, 1+3) = 4	[Index 1:3, Max Reach:4]
3	Next iteration	2	4	2 <= 4 (True)	Update max_reach = max(4, 2+1) = 4	[Index 2:1, Max Reach:4]
4	Next iteration	3	4	3 <= 4 (True)	Update max_reach = max(4, 3+1) = 4	[Index 3:1, Max Reach:4]
5	Next iteration	4	4	4 <= 4 (True)	Update max_reach = max(4, 4+4) = 8	[Index 4:4, Max Reach:8]
6	End loop	-	-	-	Reached or passed last index, return True	[Reached end, success]

💡 Loop ends because current index reached last index and max_reach covers or passes it.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	Final
i	-	0	1	2	3	4	-
max_reach	0	2	4	4	4	8	8

Key Moments - 3 Insights

Why do we check if current index i is greater than max_reach?

If i > max_reach, it means we cannot reach index i from any previous jumps, so we get stuck and must return False. This is shown in execution_table row 2 where condition i <= max_reach is checked.

Why do we update max_reach with max(max_reach, i + nums[i])?

Because from index i, we can jump up to nums[i] steps, so i + nums[i] is the farthest reachable index from i. We keep max_reach as the farthest index reachable so far. This is updated each step as shown in execution_table rows 1 to 5.

What does it mean when max_reach becomes greater than or equal to the last index?

It means we can reach or pass the last index, so the answer is True. This is the exit condition shown in execution_table row 6.

Visual Quiz - 3 Questions

Look at the execution_table at Step 3, what is the value of max_reach?

A 3

B 2

C 4

D 1

💡 Hint

Check the 'Max Reach' column at Step 3 in execution_table.

At which step does the condition i <= max_reach become false, causing the loop to stop?

A It never becomes false in this example

B Step 2

C Step 5

D Step 4

💡 Hint

Look at the 'Condition i <= Max Reach?' column in execution_table for all steps.

If nums[1] was 1 instead of 3, how would max_reach change at Step 2?

A max_reach would be 4

B max_reach would be 2

C max_reach would be 3

D max_reach would be 1

💡 Hint

Calculate max(max_reach at Step 1, i + nums[i]) with nums[1] = 1.

Concept Snapshot

Jump Game Problem:
- Start at index 0
- Track max reachable index
- For each index i, if i > max_reach, return False
- Update max_reach = max(max_reach, i + nums[i])
- If max_reach >= last index, return True
- Time: O(n), Space: O(1)

Full Transcript

The Jump Game problem asks if you can jump from the start to the end of an array where each element tells you the max jump length from that position. We start at index 0 and keep track of the farthest index we can reach, called max_reach. For each index i, if i is beyond max_reach, it means we cannot get there, so we return False. Otherwise, we update max_reach to the maximum of its current value and i plus the jump length at i. If at any point max_reach reaches or passes the last index, we return True. This approach runs in linear time and uses constant space.

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