Jump Game II (Min Jumps)

▶

Steps

setup

Initialize variables

Set jumps, current_end, and furthest to 0 before starting iteration over the array.

💡 Initialization sets the starting point: no jumps made, and boundaries at the start index.

Line:

jumps = 0
current_end = 0
furthest = 0

💡 All tracking variables start at zero, representing the initial position before any jumps.

fill_cells

Start iteration at index 0

At index 0, calculate furthest reachable index as max(0, 0 + nums[0]) = 2.

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

Line:

furthest = max(furthest, i + nums[i])  # i=0

💡 From index 0, we can jump up to index 2, updating furthest to 2.

compare

Check if at current_end at index 0

Since i == current_end (0 == 0), increment jumps to 1 and update current_end to furthest (2).

💡 Reaching current_end means we must make a jump to continue.

Line:

if i == current_end:
    jumps += 1
    current_end = furthest

💡 We commit to a jump here, extending our boundary to index 2.

fill_cells

Move to index 1

At index 1, update furthest to max(2, 1 + nums[1]) = max(2, 4) = 4.

💡 We check if we can jump further from this new position.

Line:

furthest = max(furthest, i + nums[i])  # i=1

💡 From index 1, we can jump as far as index 4, extending furthest.

compare

Check if at current_end at index 1

Index 1 is not equal to current_end (1 != 2), so no jump is made.

💡 We only jump when reaching the current boundary.

Line:

if i == current_end:
    jumps += 1
    current_end = furthest  # i=1 no jump

💡 We continue exploring without incrementing jumps.

fill_cells

Move to index 2

At index 2, update furthest to max(4, 2 + nums[2]) = max(4, 3) = 4 (no change).

💡 Check if this position extends the furthest reachable index.

Line:

furthest = max(furthest, i + nums[i])  # i=2

💡 No extension of furthest; it remains 4.

compare

Check if at current_end at index 2

Index 2 equals current_end (2 == 2), so increment jumps to 2 and update current_end to furthest (4).

💡 Reached boundary again, so we must jump to continue.

Line:

if i == current_end:
    jumps += 1
    current_end = furthest  # i=2 jump

💡 Jump count increases to 2, extending boundary to the last index.

fill_cells

Move to index 3

At index 3, update furthest to max(4, 3 + nums[3]) = max(4, 4) = 4 (no change).

💡 Check if this position extends the furthest reachable index.

Line:

furthest = max(furthest, i + nums[i])  # i=3

💡 Furthest remains 4, no extension.

compare

Check if at current_end at index 3

Index 3 is not equal to current_end (3 != 4), so no jump is made.

💡 We only jump when reaching the current boundary.

Line:

if i == current_end:
    jumps += 1
    current_end = furthest  # i=3 no jump

💡 Continue exploring without incrementing jumps.

traverse

End iteration and return jumps

Iteration ends at index n-2=3. Return jumps = 2 as minimum jumps to reach last index.

💡 The final jump count represents the minimum jumps needed.

Line:

return jumps

💡 The algorithm successfully computed the minimum jumps as 2.

def jump(nums):
    n = len(nums)
    jumps = 0  # STEP 1
    current_end = 0  # STEP 1
    furthest = 0  # STEP 1
    for i in range(n - 1):
        furthest = max(furthest, i + nums[i])  # STEP 2,4,6,8
        if i == current_end:  # STEP 3,5,7,9
            jumps += 1
            current_end = furthest
    return jumps  # STEP 10

if __name__ == '__main__':
    print(jump([2,3,1,1,4]))  # Output: 2

📊

Jump Game II (Min Jumps) - Watch the Algorithm Execute, Step by Step

Watching each update to furthest and current_end reveals how the greedy approach efficiently finds the minimum jumps without exploring all paths.

Item 0 - wt:2 val:0

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	2	?	?	?	?

furthest updated

Item 0 - wt:2 val:2

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

jump made

Item 1 - wt:3 val:4

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

furthest updated

Item 1 - wt:3 val:4

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

no jump

Item 2 - wt:1 val:4

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

furthest checked

Item 2 - wt:1 val:4

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

jump made

Item 3 - wt:1 val:4

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

furthest checked

Item 3 - wt:1 val:4

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

no jump

Item 4 - wt:4 val:4

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

answer

Key Takeaways

✓ The algorithm uses a greedy approach to track the furthest reachable index and only increments jumps when the current boundary is reached.

This insight is hard to see from code alone because the jump increments are tied to boundary conditions, not every index.

✓ Furthest reachable index monotonically increases or stays the same, never decreasing, which ensures progress towards the end.

Visualizing furthest updates clarifies how the algorithm efficiently skips unnecessary checks.

✓ Jumps increment exactly when the iteration index reaches the current_end boundary, showing the minimal jump points.

Seeing this decision in action helps understand why jumps are minimal and not overcounted.

Practice

1. You are given an array representing daily stock prices. You want to find the maximum profit possible by buying once and selling once after the buy day. Which approach guarantees the optimal solution with linear time and constant space complexity?

easy

A. Use nested loops to check all pairs of buy and sell days and track the maximum profit.

B. Use a dynamic programming table storing maximum profit for each day and each possible transaction count.

C. Iterate once, tracking the minimum price so far and updating the maximum profit by comparing current price minus minimum price.

D. Apply a greedy algorithm that always sells on the next day if the price is higher than the buy day.

Solution

1. Step 1: Understand problem constraints

   The problem requires exactly one buy and one sell, with the sell day after the buy day, maximizing profit.

2. Step 2: Identify optimal approach

   Tracking the minimum price so far and updating max profit in one pass ensures O(n) time and O(1) space, guaranteeing optimality.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   One pass with min price tracking is classic and efficient [OK]

Hint: Track min price and max profit in one pass [OK]

Common Mistakes:

• Using nested loops causes TLE on large inputs
• Greedy selling next day misses better future profits
• DP table wastes space for this simple case

2. Consider the following Python code for maximizing stock profit with transaction fee:

def maxProfit(prices, fee):
    cash, hold = 0, -prices[0]
    for price in prices[1:]:
        cash = max(cash, hold + price - fee)
        hold = max(hold, cash - price)
    return cash

prices = [1, 3, 2]
fee = 1
print(maxProfit(prices, fee))

What is the output of this code?

easy

A. 1

B. 0

C. 3

D. 2

Solution

1. Step 1: Initialize variables

   cash = 0, hold = -1 (buy at price 1)

2. Step 2: Iterate over prices

   Price=3: cash = max(0, -1 + 3 - 1) = max(0,1) = 1; hold = max(-1, 1 - 3) = max(-1,-2) = -1 Price=2: cash = max(1, -1 + 2 - 1) = max(1,0) = 1; hold = max(-1, 1 - 2) = max(-1,-1) = -1

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Final cash is 1, representing max profit after fee [OK]

Hint: Track cash and hold updates carefully [OK]

Common Mistakes:

• Off-by-one in loop
• Forgetting fee subtraction

3. You are given a string s and a dictionary of words. You need to determine if s can be segmented into a space-separated sequence of one or more dictionary words. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Pure brute force recursion checking all possible segmentations without memoization

B. Greedy approach that tries to match the longest prefix word repeatedly

C. Dynamic Programming that builds up a boolean array indicating segmentability up to each index

D. Sorting the dictionary and using binary search to find words in the string

Solution

1. Step 1: Understand the problem constraints

   The problem requires checking if the string can be segmented into dictionary words, which involves overlapping subproblems and optimal substructure.

2. Step 2: Identify the suitable algorithmic pattern

   Dynamic Programming (DP) is appropriate because it stores results of subproblems (segmentability up to index i) to avoid recomputation, guaranteeing an optimal solution.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   DP uses boolean array to track valid breaks [OK]

Hint: DP with boolean array tracks segmentability [OK]

Common Mistakes:

• Thinking greedy longest prefix always works
• Assuming brute force is efficient enough
• Using binary search without DP

4. Suppose the Jump Game problem is modified so that you can reuse positions multiple times (i.e., cycles allowed). Which approach correctly handles this variant to determine if the last index is reachable?

hard

A. Use the greedy approach as is, since it already tracks max reachable index

B. Use DFS with memoization to detect cycles and avoid infinite loops

C. Use bottom-up DP iterating forwards to accumulate reachable states

D. Use a simple BFS traversal marking visited indices to detect reachability

Solution

1. Step 1: Understand reuse implies cycles

   Allowing reuse means cycles can occur, so greedy approach may fail as it assumes forward progress.

2. Step 2: Identify correct approach

   BFS with visited marking detects cycles and explores reachable indices without infinite loops.

3. Step 3: Why other options fail

   DFS with memoization can work but is less efficient and more complex; greedy fails; bottom-up DP forwards is incorrect due to dependency issues.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   BFS handles cycles and reachability correctly [OK]

Hint: Cycles require graph traversal with visited checks [OK]

Common Mistakes:

• Assuming greedy works with cycles
• Using DFS without cycle detection
• Using bottom-up DP forwards incorrectly

5. Suppose the Word Break problem is modified so that dictionary words can be reused unlimited times in the segmentation. Which modification to the bottom-up DP approach correctly handles this variant?

hard

A. Change dp array to store counts of ways to segment instead of booleans

B. No change needed; the original DP already supports unlimited reuse of words

C. Remove the break statement after setting dp[i] = True to consider all segmentations

D. Use a recursive backtracking approach without memoization to explore all reuse possibilities

Solution

1. Step 1: Understand the reuse requirement

   Unlimited reuse means counting all possible segmentations, not just checking existence.

2. Step 2: Modify DP to count ways

   Change dp from boolean to integer counts, summing counts from dp[i - length] when substring is in dictionary.

3. Step 3: Evaluate other options

   Original DP only checks existence (boolean), removing break doesn't count ways correctly, recursion without memoization is inefficient.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Counting ways requires dp to store counts, not booleans [OK]

Hint: Counting segmentations requires dp to store counts, not booleans [OK]

Common Mistakes:

• Assuming original DP counts ways
• Removing break to count ways incorrectly
• Using naive recursion without memoization