Maximum Points You Can Obtain from Cards

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Steps

setup

Calculate total sum of all cards

We sum all the card points to get the total points available before selecting any cards.

💡 Knowing the total sum is essential because the final answer depends on subtracting the minimum subarray sum from this total.

Line:

total = sum(cardPoints)

💡 The total sum represents the maximum points if we could take all cards.

setup

Initialize sliding window size and first window sum

We set the window size to n-k (7-3=4) and calculate the sum of the first window of this size at the start of the array.

💡 This initial window sum is our starting point to find the minimum sum subarray of size n-k.

Line:

window_size = n - k
current_sum = sum(cardPoints[:window_size])
min_subarray_sum = current_sum

💡 The first window sum is the baseline minimum subarray sum to compare against as we slide the window.

expand

Slide window right by 1: include cardPoints[4], exclude cardPoints[0]

We move the window one step right by adding cardPoints[4] (5) and removing cardPoints[0] (1) from the current sum.

💡 Sliding the window updates the sum efficiently without recomputing from scratch.

Line:

current_sum += cardPoints[i] - cardPoints[i - window_size]

💡 This step shows how the sliding window maintains a fixed size and updates sums incrementally.

compare

Compare current sum with min_subarray_sum

We check if the new current sum (14) is less than the recorded minimum subarray sum (10).

💡 This comparison determines if we found a new minimum sum subarray.

Line:

if current_sum < min_subarray_sum:
    min_subarray_sum = current_sum

💡 The minimum sum remains 10 because the new window sum is larger.

expand

Slide window right by 1: include cardPoints[5], exclude cardPoints[1]

We move the window right again by adding cardPoints[5] (6) and removing cardPoints[1] (2) from the sum.

💡 Continuing to slide the window helps find the minimum sum subarray across the array.

Line:

current_sum += cardPoints[i] - cardPoints[i - window_size]

💡 The window sum updates to reflect the new subarray of cards.

compare

Compare current sum with min_subarray_sum

We check if the current sum (18) is less than the minimum subarray sum (10).

💡 This comparison decides if the minimum sum should be updated.

Line:

if current_sum < min_subarray_sum:
    min_subarray_sum = current_sum

💡 The minimum sum remains unchanged because the current window sum is larger.

expand

Slide window right by 1: include cardPoints[6], exclude cardPoints[2]

We slide the window right once more by adding cardPoints[6] (1) and removing cardPoints[2] (3) from the sum.

💡 This final slide checks the last possible subarray of size n-k.

Line:

current_sum += cardPoints[i] - cardPoints[i - window_size]

💡 The window sum updates to the last subarray sum to consider.

compare

Compare current sum with min_subarray_sum

We check if the current sum (16) is less than the minimum subarray sum (10).

💡 This comparison determines if the minimum sum should be updated one last time.

Line:

if current_sum < min_subarray_sum:
    min_subarray_sum = current_sum

💡 The minimum sum remains 10, the smallest subarray sum found.

prune

Calculate maximum points by subtracting min_subarray_sum from total

We subtract the minimum subarray sum (10) from the total sum (22) to get the maximum points obtainable (12).

💡 This subtraction gives the maximum points by excluding the minimum sum subarray.

Line:

return total - min_subarray_sum

💡 The maximum points come from taking cards outside the minimum sum subarray.

reconstruct

Return final answer

The algorithm returns 12 as the maximum points obtainable by taking cards from the ends.

💡 This final step confirms the result of the sliding window approach.

Line:

return total - min_subarray_sum

💡 The answer 12 matches the expected output, validating the algorithm.

def maxScore(cardPoints, k):
    n = len(cardPoints)  # STEP 1
    total = sum(cardPoints)  # STEP 1
    window_size = n - k  # STEP 2
    current_sum = sum(cardPoints[:window_size])  # STEP 2
    min_subarray_sum = current_sum  # STEP 2
    for i in range(window_size, n):  # STEP 3
        current_sum += cardPoints[i] - cardPoints[i - window_size]  # STEP 3
        if current_sum < min_subarray_sum:  # STEP 4
            min_subarray_sum = current_sum  # STEP 4
    return total - min_subarray_sum  # STEP 9 and 10

if __name__ == '__main__':
    print(maxScore([1,2,3,4,5,6,1], 3))  # Output: 12

📊

Maximum Points You Can Obtain from Cards - Watch the Algorithm Execute, Step by Step

Watching each step helps you understand how the sliding window technique efficiently finds the minimum subarray sum, which is key to solving this problem optimally.

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Key Takeaways

✓ The problem reduces to finding the minimum sum subarray of size n-k to maximize points from the ends.

This insight is hard to see from code alone because it requires understanding the complementary relationship between the subarray and the cards taken.

✓ Sliding window efficiently updates sums by adding the new element and removing the old, avoiding recomputation.

Visualizing each window move clarifies how the algorithm maintains performance and correctness.

✓ Not every window slide updates the minimum sum; comparisons ensure only the smallest sum is recorded.

Seeing each comparison step helps understand when and why the minimum sum changes.

Practice

1. The following code attempts to find the length of the longest substring without repeating characters. Which line contains a subtle bug that can cause incorrect results on some inputs?

medium

A. Line 7: left = index_map[char] + 1

B. Line 6: if char in index_map:

C. Line 5: for right, char in enumerate(s):

D. Line 9: max_len = max(max_len, right - left + 1)

Solution

1. Step 1: Identify the bug in updating left pointer

   The code updates left pointer unconditionally to index_map[char] + 1 when a duplicate is found, but does not check if this moves left backward, which can shrink the window incorrectly.

2. Step 2: Correct approach

   Update left only if index_map[char] + 1 is greater than current left to avoid moving left backward.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Missing boundary check on left pointer update causes wrong window size [OK]

Hint: Check if left pointer moves backward [OK]

Common Mistakes:

• Not verifying left pointer position before update
• Assuming duplicates always appear after left

2. What is the time complexity of the optimal sliding window solution that uses a queue to track zero positions for the Max Consecutive Ones III problem, given an input array of length n and integer k?

medium

A. O(n * k) because each zero can cause multiple window shifts

B. O(n) because each element is visited at most twice by the two pointers

C. O(n^2) because the queue operations inside the loop cause nested behavior

D. O(k) because the queue size is limited to k zeros only

Solution

1. Step 1: Identify loop behavior

   The right pointer iterates over all n elements once, and the left pointer only moves forward when zeros exceed k.

2. Step 2: Analyze queue operations and pointer moves

   Each zero index is enqueued once and dequeued once, so total operations are O(n). However, each zero can cause the left pointer to move multiple times in worst case, leading to O(n * k) complexity.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   While the queue operations are O(n), the window shifts can multiply by k in worst cases [OK]

Hint: Consider worst-case window shifts caused by zeros [OK]

Common Mistakes:

• Assuming strictly linear without considering multiple shifts
• Confusing queue size with total operations
• Ignoring worst-case scenarios

3. The following code attempts to find the maximum sum subarray of size k using the sliding window approach. Identify the line that contains a subtle bug that may cause incorrect results when the array contains all negative numbers.

def max_sum_subarray_sliding_window(arr, k):
    n = len(arr)
    window_sum = sum(arr[:k])
    max_sum = 0
    for i in range(k, n):
        window_sum += arr[i] - arr[i - k]
        if window_sum > max_sum:
            max_sum = window_sum
    return max_sum

medium

A. Line 4: Initializing max_sum to 0 instead of a smaller value.

B. Line 5: Loop starting from index k instead of k-1.

C. Line 3: Initializing window_sum with sum of first k elements.

D. Line 6: Updating window_sum by adding arr[i] and subtracting arr[i - k].

Solution

1. Step 1: Understand initialization of max_sum

   max_sum is initialized to 0, which is incorrect if all array elements are negative because the maximum sum could be less than 0.

2. Step 2: Consequence of wrong initialization

   If all sums are negative, max_sum remains 0, which is incorrect. Proper initialization should be max_sum = window_sum (sum of first k elements) or float('-inf').

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   max_sum initialized to 0 causes incorrect max for negative arrays [OK]

Hint: Initialize max_sum to first window sum, not zero [OK]

Common Mistakes:

• Initializing max_sum to zero instead of first window sum
• Off-by-one errors in loop indices
• Incorrect window_sum updates

4. Suppose you want to find the longest substring with at most k distinct characters, but now characters can be reused infinitely many times and the string is extremely long. Which modification to the sliding window approach is necessary to handle this efficiently?

hard

A. Use a frequency map with counts and shrink window when counts exceed k distinct characters, ignoring last seen positions

B. Maintain an ordered dictionary of last seen indices and shrink window as usual; no change needed

C. Use a frequency map and shrink the window only when distinct characters exceed k, ignoring last seen indices

D. Switch to a brute force approach since reuse breaks sliding window assumptions

Solution

1. Step 1: Understand reuse impact

   When characters can be reused infinitely, tracking last seen indices is insufficient; frequency counts are needed.

2. Step 2: Modify approach

   Use a frequency map to count occurrences and shrink window when distinct characters exceed k, ignoring last seen positions.

3. Step 3: Confirm efficiency

   This approach maintains O(n) time and handles reuse correctly.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Frequency map with counts is required for reuse scenarios [OK]

Hint: Frequency counts needed when characters can be reused infinitely [OK]

Common Mistakes:

• Assuming last seen indices suffice for reuse
• Thinking brute force is necessary

5. Suppose the sliding window maximum problem is modified so that elements can be reused multiple times (i.e., the window can slide but elements may appear multiple times in the output). Which of the following statements about the monotonic deque approach is true?

hard

A. The monotonic deque approach fails because it assumes each element appears only once; a max-heap is required instead.

B. The monotonic deque approach must be modified to track counts of elements to handle reuse correctly.

C. The monotonic deque approach works without any changes since it only depends on indices.

D. The problem reduces to a different pattern; monotonic deque is not applicable and a segment tree is needed.

Solution

1. Step 1: Understand the reuse modification

   Allowing elements to be reused means the same element can contribute multiple times as the window slides, potentially overlapping.

2. Step 2: Analyze monotonic deque assumptions

   The original deque stores indices assuming each element appears once per position. To handle reuse, counts or frequency tracking is needed to maintain correctness.

3. Step 3: Evaluate other options

   Monotonic deque without modification fails; max-heap alone is inefficient; segment tree is overkill for this variant.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Tracking counts allows correct max maintenance with reuse [OK]

Hint: Reuse breaks unique index assumption; track counts to fix [OK]

Common Mistakes:

• Assuming no changes needed for reuse
• Switching to max-heap without efficiency gain
• Thinking monotonic deque is inapplicable