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Valid Palindrome

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Steps
setup

Clean the input string

Extract only alphanumeric characters from the input and convert them to lowercase, forming the cleaned array.

💡 Cleaning simplifies the problem by ignoring spaces, punctuation, and case differences, focusing only on relevant characters.
Line:cleaned = [ch.lower() for ch in s if ch.isalnum()]
💡 The cleaned array is the true data structure the palindrome check operates on.
📊
Valid Palindrome - Watch the Algorithm Execute, Step by Step
Watching each pointer move and compare reveals how the algorithm efficiently checks palindrome validity without extra space or complex operations.
Step 1/13
·Active fill★Answer cell
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Result: true

Key Takeaways

✓ Cleaning the input string to lowercase alphanumeric characters is essential before palindrome checking.

Without cleaning, spaces and punctuation would cause false mismatches, making the algorithm incorrect.

✓ Two pointers moving inward from opposite ends efficiently check palindrome property without extra space.

This approach avoids reversing the string or using extra memory, making it optimal.

✓ If any character pair mismatches, the algorithm immediately returns false, stopping unnecessary checks.

Early termination saves time and shows the power of the two-pointer technique.

Practice

(1/5)
1. You are given an integer array and asked to find all unique triplets that sum to zero. Which approach guarantees an optimal solution in terms of time complexity and handles duplicates efficiently?
easy
A. Sort the array, fix one element, and use a two-pointer approach to find pairs that sum to the negative of the fixed element, skipping duplicates.
B. Use a greedy approach by sorting and picking the smallest and largest elements to form triplets.
C. Use dynamic programming to build up solutions for smaller subsets and combine them to find triplets.
D. Use three nested loops to check all triplets and store unique sorted triplets in a set.

Solution

  1. Step 1: Understand problem constraints

    We need all unique triplets summing to zero, efficiently handling duplicates.

  2. Step 2: Identify approach that handles duplicates and optimizes search

    Sorting and fixing one element allows using a two-pointer approach to find pairs summing to the complement, skipping duplicates efficiently.

  3. Final Answer:

    Option A -> Option A

  4. Quick Check:

    Sorting + two-pointer approach is O(n^2) and handles duplicates [OK]
Hint: Sorting + two-pointer handles duplicates and runs in O(n²) [OK]
Common Mistakes:
  • Thinking brute force is optimal
  • Using DP which doesn't fit this problem
  • Greedy approach misses some triplets
2. Consider the following Python code that counts pairs with absolute difference at most K using binary search. What is the final output when the input is nums = [1, 2, 5] and K = 3? 
import bisect

def count_pairs_binary_search(nums, K):
    count = 0
    n = len(nums)
    for i in range(n - 1):
        target = nums[i] + K
        j = bisect.bisect_right(nums, target, i + 1) - 1
        if j > i:
            count += j - i
    return count

print(count_pairs_binary_search([1, 2, 5], 3))
easy
A. 2
B. 3
C. 4
D. 5

Solution

  1. Step 1: Trace iteration i=0

    nums[0] = 1, target = 1 + 3 = 4. bisect_right(nums, 4, 1) returns index 2 (nums[2]=5 >4), so j = 2 - 1 = 1. Count += 1 - 0 = 1.

  2. Step 2: Trace iteration i=1

    nums[1] = 2, target = 2 + 3 = 5. bisect_right(nums, 5, 2) returns 3 (end of list), so j = 3 - 1 = 2. Count += 2 - 1 = 1. Total count = 2.

  3. Final Answer:

    Option A -> Option A

  4. Quick Check:

    Pairs are (1,2) and (2,5) with differences 1 and 3 respectively [OK]
Hint: Check bisect_right returns and subtract 1 carefully [OK]
Common Mistakes:
  • Off-by-one error in bisect index adjustment
  • Counting pairs where i >= j
  • Miscomputing target as nums[i] - K
3. Suppose the problem changes: you want to count pairs with absolute difference at most K, but now the array can contain duplicate elements and you want to count pairs where elements can be reused multiple times (i.e., pairs (i, j) with i ≤ j). Which modification to the binary search approach correctly handles this variant?
hard
A. Start binary search from i instead of i + 1 and count pairs where j ≥ i, adjusting count accordingly.
B. Use brute force nested loops to count all pairs with i ≤ j and difference ≤ K, as binary search cannot handle reuse.
C. Use two pointers with left starting at 0 and right starting at 0, moving right pointer forward and counting pairs including i = j.
D. Keep the same code but change bisect_right to bisect_left to include duplicates and allow i ≤ j pairs.

Solution

  1. Step 1: Understand reuse and duplicates impact

    Allowing i ≤ j means pairs with the same element twice are valid; binary search must include current index.

  2. Step 2: Modify binary search start index

    Starting bisect_right from i (not i+1) includes pairs where i == j; counting j ≥ i pairs correctly.

  3. Step 3: Adjust counting logic

    Count pairs as j - i + 1 if including i == j, or carefully count pairs with j ≥ i.

  4. Final Answer:

    Option A -> Option A

  5. Quick Check:

    Starting search at i and counting pairs with j ≥ i handles reuse correctly [OK]
Hint: Start binary search at i to include pairs with i == j [OK]
Common Mistakes:
  • Using bisect_left instead of bisect_right incorrectly
  • Ignoring pairs where i == j
  • Assuming two pointers cannot handle reuse
4. Suppose the problem changes: you want to find the minimum difference between k closest elements, but now elements can be negative and the array is very large. Which modification to the optimal approach is necessary to handle this scenario efficiently?
hard
A. Use the same sorting + sliding window approach; negative values do not affect correctness.
B. Use a balanced binary search tree to maintain a sliding window of size k for efficient min and max retrieval without sorting.
C. Use a brute force approach since negative values break the sliding window logic.
D. Sort the array and use a heap to track minimum differences instead of sliding window.

Solution

  1. Step 1: Consider effect of negative values

    Sorting works correctly regardless of negative or positive values.

  2. Step 2: Confirm sliding window validity

    Sliding window over sorted array still finds minimal difference subsets efficiently.

  3. Step 3: Evaluate other options

    Balanced BST or heap add unnecessary complexity; brute force is inefficient.

  4. Final Answer:

    Option A -> Option A

  5. Quick Check:

    Sorting + sliding window handles negatives naturally [OK]
Hint: Sorting handles negatives; sliding window logic unchanged [OK]
Common Mistakes:
  • Assuming negatives break sliding window approach
  • Overcomplicating with data structures
  • Switching to brute force unnecessarily
5. If the problem changes so that the same element can be used multiple times to reach the target sum (i.e., elements can be reused), which approach correctly solves the problem?
hard
A. Use the two pointers approach moving inward, but allow pointers to stay if sum < target to reuse elements.
B. Use a modified binary search that allows searching from the current index (not i+1) to reuse elements.
C. Use a dynamic programming approach to find combinations allowing repeated elements.
D. Use the brute force nested loops but break early when sum exceeds target.

Solution

  1. Step 1: Understand reuse requirement

    Allowing reuse means pairs can include the same element multiple times, which breaks two-pointer and binary search assumptions.

  2. Step 2: Identify suitable approach

    Dynamic programming can handle combinations with repeated elements efficiently, unlike two pointers or binary search.

  3. Final Answer:

    Option C -> Option C

  4. Quick Check:

    DP handles repeated elements and target sums correctly [OK]
Hint: Reuse -> DP for combinations, not two pointers or binary search [OK]
Common Mistakes:
  • Trying to reuse pointers incorrectly
  • Modifying binary search without correctness
  • Assuming brute force is efficient
Recommended

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10 questions~5 mineasy hard