Minimum Consecutive Cards to Pick Up

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Steps

setup

Initialize pointers and data structures

We start with both left and right pointers at index 0, an empty map to track last seen indices, and the minimum length set to infinity.

💡 Initialization sets the stage for scanning the array and tracking duplicates efficiently.

Line:

last_seen = {}
left = 0
min_len = float('inf')

💡 We have a clean slate with no cards processed yet and no duplicates found.

expand

Process first card (3) at right=0

We process the card 3 at index 0. Since it is not in last_seen, no left pointer movement is needed. We record last_seen[3] = 0.

💡 Recording last seen indices helps detect duplicates later.

Line:

if card in last_seen and last_seen[card] >= left:
    left = last_seen[card] + 1
last_seen[card] = right

💡 The window currently contains a unique card, so no shrinking is needed.

expand

Expand right pointer to 1 (card 4)

We move right pointer to index 1 and process card 4. It is not in last_seen, so no left pointer change. Update last_seen[4] = 1.

💡 Expanding the window includes new cards and checks for duplicates.

Line:

if card in last_seen and last_seen[card] >= left:
    left = last_seen[card] + 1
last_seen[card] = right

💡 Window now contains [3,4], both unique cards.

expand

Expand right pointer to 2 (card 2)

Right pointer moves to index 2, card 2 is new. No duplicate inside window, update last_seen[2] = 2.

💡 Continuing to expand the window while cards are unique.

Line:

if card in last_seen and last_seen[card] >= left:
    left = last_seen[card] + 1
last_seen[card] = right

💡 Window now contains [3,4,2], all unique.

compare

Expand right pointer to 3 (card 3) - duplicate found

Right pointer moves to index 3, card 3 was last seen at index 0, which is inside current window (left=0). We must move left pointer to 1 (0+1) to exclude previous 3.

💡 Detecting duplicates triggers shrinking the window from the left to maintain uniqueness.

Line:

if card in last_seen and last_seen[card] >= left:
    left = last_seen[card] + 1

💡 Left pointer moves to exclude the first occurrence of 3, shrinking the window.

insert

Update last_seen for card 3 at index 3

We update last_seen[3] = 3 to record the latest occurrence of card 3.

💡 Updating last seen index is crucial to detect future duplicates correctly.

Line:

last_seen[card] = right

💡 The map now reflects the most recent position of card 3.

record_answer

Update minimum length with current window size

Current window size is right - left + 1 = 3 - 1 + 1 = 3. Since min_len was infinity, update min_len = 3.

💡 We record the smallest window length containing duplicates found so far.

Line:

if right - left + 1 > 1:
    min_len = min(min_len, right - left + 1)

💡 We have found a candidate minimum subarray length containing duplicates.

compare

Expand right pointer to 4 (card 4) - duplicate found

Right pointer moves to index 4, card 4 last seen at index 1, which is inside current window (left=1). Move left pointer to 2 (1+1) to exclude previous 4.

💡 Another duplicate triggers window shrinking from the left.

Line:

if card in last_seen and last_seen[card] >= left:
    left = last_seen[card] + 1

💡 Window adjusts to exclude the first occurrence of card 4.

insert

Update last_seen for card 4 at index 4

Update last_seen[4] = 4 to record the latest occurrence of card 4.

💡 Keeping last_seen updated ensures correct duplicate detection.

Line:

last_seen[card] = right

💡 The map now reflects the latest position of card 4.

record_answer

Update minimum length with current window size

Current window size is right - left + 1 = 4 - 2 + 1 = 3. min_len is currently 3, so it remains unchanged.

💡 We only update min_len if the current window is smaller.

Line:

if right - left + 1 > 1:
    min_len = min(min_len, right - left + 1)

💡 The minimum length remains 3 after this step.

expand

Expand right pointer to 5 (card 7) - no duplicate

Right pointer moves to index 5, card 7 is new. No duplicate inside window, left pointer stays at 2. Update last_seen[7] = 5.

💡 Expanding window with unique card does not change left pointer.

Line:

if card in last_seen and last_seen[card] >= left:
    left = last_seen[card] + 1
last_seen[card] = right

💡 Window now contains [2,3,4,7].

record_answer

Update minimum length with current window size

Current window size is 5 - 2 + 1 = 4. Since 4 > 1 and min_len is 3, update min_len = min(3,4) = 3 (no change).

💡 We check if the current window is a better candidate for minimum length.

Line:

if right - left + 1 > 1:
    min_len = min(min_len, right - left + 1)

💡 Minimum length remains 3, but window size is larger.

reconstruct

End of array reached, finalize answer

We have processed all cards. The minimum length found is 3, which corresponds to the smallest subarray with duplicate cards, such as [4,2,3] from indices 1 to 3.

💡 Final step returns the minimum length or -1 if none found.

Line:

return min_len if min_len != float('inf') else -1

💡 The algorithm correctly returns 3 as the smallest subarray length containing duplicates.

def minimumCardPickup(cards):
    last_seen = {}  # STEP 1
    left = 0       # STEP 1
    min_len = float('inf')  # STEP 1
    for right, card in enumerate(cards):  # STEP 2-13
        if card in last_seen and last_seen[card] >= left:  # STEP 5,8
            left = last_seen[card] + 1  # STEP 5,8
        last_seen[card] = right  # STEP 6,9
        if right - left + 1 > 1:  # STEP 7,10,12
            min_len = min(min_len, right - left + 1)  # STEP 7,10,12
    return min_len if min_len != float('inf') else -1  # STEP 13

if __name__ == '__main__':
    print(minimumCardPickup([3,4,2,3,4,7]))  # Output: 3
    print(minimumCardPickup([1,0,5,3]))      # Output: -1

📊

Minimum Consecutive Cards to Pick Up - Watch the Algorithm Execute, Step by Step

Watching the algorithm step-by-step reveals how the window boundaries move to exclude duplicates and how the minimum subarray length is tracked, making the logic clear without needing to read code.

expand

L=R

3

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Window: [3]

expand

L=R

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Window: [3]

expand

L ▼

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Window: [3, 4]

expand

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Window: [3, 4, 2]

shrink

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Window: [4, 2, 3]

Metric: 3

insert

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Window: [4, 2, 3]

Metric: 3

record_answer

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Window: [4, 2, 3]

Metric: 3

Answer: 3

shrink

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Window: [2, 3, 4]

Metric: 3

Answer: 3

insert

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Window: [2, 3, 4]

Metric: 3

Answer: 3

record_answer

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Window: [2, 3, 4]

Metric: 3

Answer: 3

expand

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Window: [2, 3, 4, 7]

Metric: 4

Answer: 3

record_answer

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Window: [2, 3, 4, 7]

Metric: 4

Answer: 3

record_answer

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Window: [4, 2, 3]

Metric: 3

Answer: 3

Key Takeaways

✓ The sliding window moves dynamically to exclude duplicates by adjusting the left pointer based on last seen indices.

This dynamic adjustment is hard to grasp from code alone but becomes clear when watching the window boundaries move step-by-step.

✓ The last_seen map is essential for quickly detecting duplicates inside the current window and deciding when to shrink.

Visualizing last_seen updates alongside pointer moves helps understand how duplicates are tracked efficiently.

✓ The minimum length is updated only when duplicates are found, showing how the algorithm focuses on valid subarrays.

Seeing the answer update only on duplicates clarifies why some window expansions do not affect the result.

Practice

1. Consider the following Python code snippet implementing the optimal sliding window approach to count subarrays with at most k odd numbers. What is the final return value when calling countNiceSubarrays([1,2,1], 2)?

def countAtMost(nums, k):
    count = 0
    left = 0
    odd_count = 0
    for right in range(len(nums)):
        if nums[right] % 2 == 1:
            odd_count += 1
        while odd_count > k:
            if nums[left] % 2 == 1:
                odd_count -= 1
            left += 1
        count += right - left + 1
    return count

def countNiceSubarrays(nums, k):
    return countAtMost(nums, k) - countAtMost(nums, k - 1)

easy

A. 3

B. 1

C. 2

D. 4

Solution

1. Step 1: Calculate countAtMost(nums, 2)

   Iterate right pointer: - right=0 (1 odd): count=1 - right=1 (1 odd): count=1+(1-0+1)=3 - right=2 (2 odds): count=3+(2-0+1)=6 So countAtMost(nums,2)=6

2. Step 2: Calculate countAtMost(nums, 1)

   Iterate right pointer: - right=0 (1 odd): count=1 - right=1 (1 odd): count=1+(1-0+1)=3 - right=2 (2 odds >1): move left to 1, odd_count=1, count=3+(2-1+1)=5 So countAtMost(nums,1)=5 Finally, countNiceSubarrays=6-5=1

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Manual trace confirms result 1 [OK]

Hint: Subtract countAtMost(k-1) from countAtMost(k) [OK]

Common Mistakes:

• Off-by-one in counting subarrays
• Miscounting odd numbers during window shrink
• Confusing countAtMost with exact count

2. What is the time complexity of the optimized sliding window algorithm that uses filtered characters for the minimum window substring problem, where n is the length of s and m is the length of t?

medium

A. O(n) because each character in s is visited at most twice by the two pointers

B. O(n * m) because for each character in s we check all characters in t

C. O(n + m) because we filter s and then slide over filtered characters

D. O(n^2) because shrinking and expanding the window can cause nested loops

Solution

1. Step 1: Identify main loops

   The algorithm uses two pointers moving over filtered characters, each pointer advances at most n times.

2. Step 2: Analyze filtering and counting

   Filtering s takes O(n), sliding window over filtered characters is O(n) since each character is visited twice max.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Two-pointer sliding window is linear time [OK]

Hint: Two pointers each move forward at most n times [OK]

Common Mistakes:

• Confusing filtering cost
• Assuming nested loops cause O(n^2)
• Ignoring that each pointer moves only forward

3. Suppose the fruits problem is modified so that you can pick fruits from trees non-contiguously but still want to maximize the total number of fruits collected with at most two fruit types. Which approach correctly solves this variant?

hard

A. Count total occurrences of each fruit type and sum the two largest counts

B. Sort fruits and use two pointers to find longest contiguous segment with two types

C. Use dynamic programming to find longest subsequence with at most two distinct fruits

D. Use the same sliding window approach since it handles at most two fruit types

Solution

1. Step 1: Understand non-contiguous picking

   Since picking is non-contiguous, order does not matter; we want max total fruits of two types.

2. Step 2: Optimal solution is sum of two largest fruit counts

   Count frequency of each fruit type, pick top two counts, sum them for max fruits.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Non-contiguous means subsequence, so frequency sums suffice [OK]

Hint: Non-contiguous -> sum top two fruit counts [OK]

Common Mistakes:

• Applying sliding window to non-contiguous
• Using DP unnecessarily
• Sorting loses subsequence property

4. Suppose the problem is modified so you can pick cards from the start or end any number of times, including reusing cards multiple times (infinite picks allowed). Which approach correctly adapts to this variant?

hard

A. Use a dynamic programming approach to find the maximum sum of k picks allowing repeated cards from ends.

B. Use prefix sums only on the first k cards from the start, ignoring the end cards.

C. Use a greedy approach that always picks the card with the highest point value from either end until k picks are made.

D. Use the original sliding window approach on the middle subarray since it still finds the minimal subarray to exclude.

Solution

1. Step 1: Understand the new constraint

   Reusing cards means picks are not limited to unique cards; the problem becomes unbounded with repetition allowed.

2. Step 2: Recognize sliding window limitation

   The sliding window approach assumes each card can be picked once; it fails with infinite reuse.

3. Step 3: Use dynamic programming

   DP can model picking cards with repetition from ends, tracking maximum sums for each number of picks.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   DP handles repeated picks; sliding window does not [OK]

Hint: Reusing cards breaks sliding window; DP needed [OK]

Common Mistakes:

• Assuming sliding window still works with repeated picks
• Greedy fails to consider future picks
• Ignoring picks from the end

5. Suppose the problem is modified so that characters in s1 can be reused multiple times in the substring of s2 (i.e., the substring can contain repeated characters beyond the counts in s1). Which modification to the sliding window approach correctly handles this variant?

hard

A. Remove the match count optimization and compare frequency maps at every window slide to allow repeated characters.

B. Keep the same sliding window with frequency counts and match count, but ignore counts exceeding those in s1.

C. Use a sliding window that only checks if all characters in s1 appear at least once in the window, ignoring exact counts.

D. Modify the algorithm to only check if the window contains all unique characters of s1 regardless of frequency.

Solution

1. Step 1: Understand the variant

   Allowing reuse means the substring can have more occurrences of characters than in s1. The previous match count optimization relies on exact frequency matches, which no longer applies.

2. Step 2: Identify correct approach

   To handle repeated characters beyond s1 counts, the algorithm must compare full frequency maps at each window slide, since match count optimization assumes exact matches.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Match count optimization breaks if counts can exceed s1 counts [OK]

Hint: Exact frequency match optimization fails if counts can exceed s1 [OK]

Common Mistakes:

• Assuming match count optimization still works
• Ignoring frequency counts and only checking presence