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DSA Pythonprogramming~10 mins

Jump Game Problem in DSA Python - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to check if we can jump from the first position.

DSA Python
def can_jump(nums):
    max_reach = 0
    for i in range(len(nums)):
        if i > [1]:
            return False
        max_reach = max(max_reach, i + nums[i])
    return True
Drag options to blanks, or click blank then click option'
Anums[i]
Bmax_reach
Ci
Dlen(nums)
Attempts:
3 left
💡 Hint
Common Mistakes
Using i instead of max_reach in the if condition.
Comparing with nums[i] instead of max_reach.
2fill in blank
medium

Complete the code to update the maximum reachable index.

DSA Python
def can_jump(nums):
    max_reach = 0
    for i in range(len(nums)):
        if i > max_reach:
            return False
        max_reach = [1](max_reach, i + nums[i])
    return True
Drag options to blanks, or click blank then click option'
Aabs
Bmin
Csum
Dmax
Attempts:
3 left
💡 Hint
Common Mistakes
Using min() instead of max() causes wrong max_reach updates.
Using sum() or abs() which are not suitable here.
3fill in blank
hard

Fix the error in the loop to iterate over all indices.

DSA Python
def can_jump(nums):
    max_reach = 0
    for i in [1](len(nums)):
        if i > max_reach:
            return False
        max_reach = max(max_reach, i + nums[i])
    return True
Drag options to blanks, or click blank then click option'
Arange
Bxrange
Cenumerate
Dlist
Attempts:
3 left
💡 Hint
Common Mistakes
Using enumerate() returns tuples, not just indices.
Using list(len(nums)) causes error.
4fill in blank
hard

Fill both blanks to create a function that returns True if last index is reachable.

DSA Python
def can_jump(nums):
    max_reach = 0
    for i in [1](len(nums)):
        if i > max_reach:
            return [2]
        max_reach = max(max_reach, i + nums[i])
    return True
Drag options to blanks, or click blank then click option'
Arange
BFalse
CTrue
Dlen
Attempts:
3 left
💡 Hint
Common Mistakes
Returning True inside the loop incorrectly.
Using len() instead of range() for iteration.
5fill in blank
hard

Fill all three blanks to create a dictionary comprehension that maps each index to its max reachable position if reachable.

DSA Python
def jump_map(nums):
    max_reach = [3]
    return {i: i + nums[i] for i in [1](len(nums)) if i [2] max_reach}
Drag options to blanks, or click blank then click option'
Arange
B<=
C0
Dlen
Attempts:
3 left
💡 Hint
Common Mistakes
Using < instead of <= causes missing reachable index.
Initializing max_reach incorrectly.

Practice

(1/5)
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
Recommended

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10 questions~5 mineasy hard