Jump Game

▶

Steps

setup

Initialize maxReach and variables

We start by setting maxReach to 0, meaning initially we can only reach the first index. We also determine the length of the input array.

💡 Initializing maxReach is crucial because it represents the furthest index reachable so far, which guides all future decisions.

Line:

maxReach = 0
n = len(nums)

💡 maxReach starts at 0 because we begin at the first index.

compare

Check if index 0 is reachable

We check if the current index (0) is greater than maxReach (0). Since it is not, we proceed.

💡 If the current index were beyond maxReach, it would mean we cannot reach this position, so the algorithm would return false.

Line:

if i > maxReach:
    return False

💡 Index 0 is reachable because it is the starting point.

fill_cells

Update maxReach at index 0

We update maxReach to the maximum of current maxReach (0) and i + nums[i] (0 + 2 = 2). maxReach becomes 2.

💡 Updating maxReach shows how far we can jump from the current position.

Line:

maxReach = max(maxReach, i + nums[i])

💡 From index 0, we can reach up to index 2.

compare

Check if maxReach reaches last index after index 0

We check if maxReach (2) is greater than or equal to last index (4). It is not, so we continue.

💡 If maxReach reached the last index, we could return true immediately.

Line:

if maxReach >= n - 1:
    return True

💡 We need to continue checking further indices.

compare

Move to index 1 and check reachability

At index 1, we check if 1 is greater than maxReach (2). It is not, so index 1 is reachable.

💡 Ensuring the current index is reachable is necessary before updating maxReach.

Line:

if i > maxReach:
    return False

💡 Index 1 is reachable based on previous maxReach.

fill_cells

Update maxReach at index 1

Update maxReach to max(2, 1 + 3) = 4. maxReach now reaches the last index.

💡 This update shows how jumps from index 1 extend our reach to the end.

Line:

maxReach = max(maxReach, i + nums[i])

💡 From index 1, we can jump directly to or beyond the last index.

compare

Check if maxReach reaches last index after index 1

Check if maxReach (4) >= last index (4). It is, so we return true.

💡 Reaching or passing the last index means the end is reachable.

Line:

if maxReach >= n - 1:
    return True

💡 The algorithm confirms the array is jumpable to the end.

reconstruct

Final state: answer is true

The algorithm returns true, indicating the last index is reachable from the start.

💡 This final step confirms the success of the greedy approach.

Line:

return True

💡 The problem is solved with a simple linear scan and maxReach updates.

traverse

Summary of maxReach progression

maxReach started at 0, updated to 2 at index 0, then to 4 at index 1, reaching the end.

💡 Seeing maxReach values side-by-side clarifies the algorithm's progress.

💡 maxReach monotonically increases or stays the same, never decreasing.

reconstruct

End of iteration and final return

Since maxReach reached the last index, the function returns true, confirming the jump game is solvable.

💡 The final return value is the culmination of all previous steps.

Line:

return True

💡 The greedy approach efficiently solves the problem in linear time.

def canJump(nums):
    maxReach = 0  # STEP 1: Initialize maxReach
    n = len(nums)  # STEP 1: Get length
    for i in range(n):  # STEP 2-10: Iterate over indices
        if i > maxReach:  # STEP 2,5: Check reachability
            return False
        maxReach = max(maxReach, i + nums[i])  # STEP 3,6: Update maxReach
        if maxReach >= n - 1:  # STEP 4,7: Check if end reachable
            return True
    return False  # STEP 10: Return false if end not reached

if __name__ == '__main__':
    print(canJump([2,3,1,1,4]))  # True
    print(canJump([3,2,1,0,4]))  # False

📊

Jump Game - Watch the Algorithm Execute, Step by Step

Watching this step-by-step helps you understand how the greedy approach efficiently determines reachability without exploring all jump paths.

Item 0 - wt:2 val:2

i\w	0	1	2	3	4
i=0	0	?	?	?	?

start

Item 0 - wt:2 val:2

i\w	0	1	2	3	4
i=0	0	?	?	?	?

reachable

Item 0 - wt:2 val:2

i\w	0	1	2	3	4
i=0	2	?	?	?	?

maxReach updated

Item 0 - wt:2 val:2

i\w	0	1	2	3	4
i=0	2	?	?	?	?

maxReach < last

Item 1 - wt:3 val:3

i\w	0	1	2	3	4
i=0	2	?	?	?	?

reachable

Item 1 - wt:3 val:3

i\w	0	1	2	3	4
i=0	2	4	?	?	?

maxReach updated

Item 1 - wt:3 val:3

i\w	0	1	2	3	4
i=0	2	4	?	?	?

goal reached

Item 1 - wt:3 val:3

i\w	0	1	2	3	4
i=0	2	4	?	?	?

answer cell

Item 1 - wt:3 val:3

i\w	0	1	2	3	4
i=0	2	4	?	?	?

maxReach=2

Item 1 - wt:3 val:3

i\w	0	1	2	3	4
i=0	2	4	?	?	?

final answer

Key Takeaways

✓ The maxReach variable tracks the furthest index reachable at each step, enabling a linear-time greedy solution.

This insight is hard to see from code alone because the variable's role is abstract without visualization.

✓ If the current index exceeds maxReach, the algorithm immediately returns false, showing unreachable positions.

Visualizing this condition clarifies why the algorithm can stop early without checking all paths.

✓ maxReach never decreases; it either stays the same or increases, reflecting monotonic growth in reachable indices.

Seeing this monotonic property helps understand why the greedy approach is correct and efficient.

Practice

1. You are given a list of daily stock prices and want to maximize profit by buying and selling any number of times, but you cannot hold more than one stock at a time. Which algorithmic approach guarantees the optimal solution for this problem?

easy

A. Greedy approach summing all positive price differences between consecutive days

B. Dynamic Programming using a state machine to track holding and not holding states

C. Pure brute force recursion exploring all buy/sell combinations without memoization

D. Divide and conquer by splitting the price array and merging results

Solution

1. Step 1: Understand problem constraints

   The problem allows unlimited transactions but only one stock can be held at a time.

2. Step 2: Identify suitable algorithm

   For this problem, the greedy approach of summing all positive price differences between consecutive days guarantees the optimal profit. This works because any profitable transaction can be decomposed into these increments.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Greedy summing positive differences is optimal for unlimited transactions without cooldown or fees [OK]

Hint: Sum all positive price differences between consecutive days to get max profit [OK]

Common Mistakes:

• Assuming DP is required when greedy suffices
• Using brute force without memoization
• Trying divide and conquer without state tracking

2. You want to find the minimum total cost to reach the top of a staircase where you can climb either one or two steps at a time, and each step has an associated cost. Which algorithmic approach guarantees finding the minimum cost efficiently?

easy

A. Greedy algorithm that always chooses the step with the lowest immediate cost

B. Pure brute force recursion exploring all possible step sequences without memoization

C. Dynamic programming that builds the solution bottom-up by storing minimum costs for each step

D. Divide and conquer by splitting the staircase into halves and solving independently

Solution

1. Step 1: Understand problem constraints and DP suitability

   The problem requires minimizing total cost with overlapping subproblems and optimal substructure, which suits dynamic programming.

2. Step 2: Evaluate algorithm suitability

   Greedy fails because local minimum cost steps do not guarantee global minimum. Brute force is correct but inefficient. Divide and conquer does not exploit overlapping subproblems. Bottom-up DP efficiently computes minimum cost for each step using previous results.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   DP stores and reuses subproblem results for optimal solution [OK]

Hint: DP bottom-up handles overlapping subproblems optimally [OK]

Common Mistakes:

• Assuming greedy approach always works
• Confusing brute force with DP
• Ignoring overlapping subproblems

3. What is the time complexity of the optimal solution that tracks the minimum price and maximum profit in one pass for the Best Time to Buy and Sell Stock (1 tx) problem?

medium

A. O(n) because it iterates through the prices once

B. O(n^2) because it compares all pairs of days

C. O(n log n) due to sorting or binary search involved

D. O(1) because it uses constant space

Solution

1. Step 1: Identify loop structure

   The algorithm iterates through the prices array exactly once, performing constant time operations per element.

2. Step 2: Confirm no nested loops or sorting

   There are no nested loops or sorting steps, so time complexity is linear in input size.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Single pass over n elements -> O(n) time [OK]

Hint: One loop over prices array -> O(n) time [OK]

Common Mistakes:

• Confusing space complexity with time complexity
• Assuming nested loops due to DP mention
• Thinking sorting is required

4. If the problem is changed to allow unlimited transactions instead of at most two, which modification to the DP approach is correct?

hard

A. Keep the same DP with k=2 but run the loop more times

B. Use a simpler DP tracking only holding and not holding states without transaction count

C. Use the same code but reset profit and maxDiff arrays after every two transactions

D. Use brute force to check all possible transaction pairs

Solution

1. Step 1: Understand unlimited transactions impact

   With unlimited transactions, tracking transaction count is unnecessary; only holding or not holding states matter.

2. Step 2: Simplify DP states

   Use a DP that tracks max profit holding stock or not holding stock each day, allowing multiple buys and sells without limit.

3. Step 3: Reject other options

   Keeping k=2 or resetting arrays is incorrect; brute force is inefficient and unnecessary.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Unlimited transactions reduce problem to simpler state machine [OK]

Hint: Unlimited transactions remove need to track transaction count [OK]

Common Mistakes:

• Trying to reuse k=2 DP for unlimited
• Resetting DP arrays incorrectly

5. Suppose the stock trading problem is modified so that you must pay a transaction fee for each sale. Which of the following changes to the state machine DP solution correctly accounts for the fee while maintaining optimal profit calculation?

hard

A. Add the fee to both buying and selling updates to ensure fee is counted twice

B. Subtract the fee from the buying price when updating holding: holding = max(holding, prev_not_holding - prices[i] - fee)

C. Subtract the fee from the selling price when updating not_holding: not_holding = max(not_holding, holding + prices[i] - fee)

D. Ignore the fee in DP and subtract total fee at the end from final profit

Solution

1. Step 1: Understand fee impact

   The fee applies on each sale, so it should reduce the profit gained when selling.

2. Step 2: Adjust DP transitions

   Subtract fee from the selling transition: not_holding = max(not_holding, holding + prices[i] - fee). Buying price remains unchanged.

3. Step 3: Why other options fail

   Subtracting fee on buying overcharges; adding fee twice double counts; ignoring fee until end breaks DP correctness.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Fee reduces profit only on selling, correctly modeled in not_holding update [OK]

Hint: Fee applies only on selling, subtract from selling transition [OK]

Common Mistakes:

• Subtracting fee on buying instead of selling
• Double counting fee on buy and sell
• Ignoring fee in DP transitions