Palindrome Number

📋

Problem

Imagine you are building a system that validates user IDs which must read the same forwards and backwards to ensure symmetry and ease of recall.

Given an integer x, determine whether it is a palindrome. An integer is a palindrome when it reads the same backward as forward. Return true if x is a palindrome, and false otherwise.

-2^31 <= x <= 2^31 - 1

Edge cases: 0 → true (single digit is always palindrome)-101 → false (negative number)1001 → true (even length palindrome)

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IDE

def is_palindrome(x: int) -> bool:public boolean isPalindrome(int x)bool isPalindrome(int x)function isPalindrome(x)

def is_palindrome(x: int) -> bool:
    # Write your solution here
    pass

class Solution {
    public boolean isPalindrome(int x) {
        // Write your solution here
        return false;
    }
}

#include <vector>
using namespace std;

bool isPalindrome(int x) {
    // Write your solution here
    return false;
}

function isPalindrome(x) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: true for negative numbersDid not check for negative input and returned true incorrectly.✅ Add condition: if x < 0: return false

Wrong: true for numbers with trailing zeros like 100Failed to handle trailing zero edge case causing false positives.✅ Return false if x % 10 == 0 and x != 0

Wrong: true for non-palindromes like 10Used greedy approach comparing only first and last digits.✅ Compare all digit pairs from both ends until pointers meet.

Wrong: false for palindrome numbers with even length like 123321Confused full reversal with half reversal approach causing incorrect result.✅ Reverse only half the digits and compare with remaining half.

Wrong: TLE on large inputsUsed full string conversion or full reversal causing O(n) or worse complexity.✅ Implement half reversal method with O(log n) complexity.

✓

Test Cases

Focus on handling edge cases like negatives, zeros, and single digits.

Beware of common pitfalls like greedy checks and full reversal overflow.

Optimize your solution to run in logarithmic time by reversing half the number.

t1_01basic

Input{"x":121}

Expectedtrue

⏱ Performance - must finish in 2000ms

Reading 121 from left to right and right to left is the same.

💡 Check if the number reads the same forwards and backwards.

💡 Try converting the number to a string and compare characters from both ends.

💡 Use two pointers moving inward to verify palindrome property.

Why it failed: Returned false for input 121 - likely failed to compare digits symmetrically. Fix by checking characters from both ends until pointers meet.

✓ Correctly identified 121 as a palindrome.

t1_02basic

Input{"x":12321}

Expectedtrue

⏱ Performance - must finish in 2000ms

12321 reads the same forwards and backwards, so it is a palindrome.

💡 Consider the symmetry of digits in the number.

💡 Check pairs of digits from start and end moving towards the center.

💡 Return false immediately if any pair mismatches; otherwise true.

Why it failed: Returned false for 12321 - likely did not check all digit pairs or stopped early. Fix by fully comparing digit pairs until center.

✓ Correctly identified 12321 as a palindrome.

t2_01edge

Input{"x":0}

Expectedtrue

⏱ Performance - must finish in 2000ms

Single digit 0 is always a palindrome.

💡 Single digit numbers are palindromes by definition.

💡 Check if input length is 1 or number is between 0 and 9.

💡 Return true immediately for single digit inputs.

Why it failed: Returned false for 0 - likely missing base case for single digit numbers. Fix by returning true if x is between 0 and 9 inclusive.

✓ Correctly handled single digit palindrome 0.

t2_02edge

Input{"x":-101}

Expectedfalse

⏱ Performance - must finish in 2000ms

Negative numbers are not palindromes due to the minus sign.

💡 Negative numbers cannot be palindromes because of the leading '-'.

💡 Check if the number is negative and return false immediately.

💡 Add a condition to return false if x < 0.

Why it failed: Returned true for -101 - failed to handle negative input. Fix by returning false if x < 0.

✓ Correctly rejected negative number -101 as palindrome.

t2_03edge

Input{"x":1001}

Expectedtrue

⏱ Performance - must finish in 2000ms

1001 reads the same forwards and backwards, even length palindrome.

💡 Check symmetry of digits even for even length numbers.

💡 Compare digit pairs from both ends moving inward.

💡 Ensure trailing zeros do not cause incorrect rejection.

Why it failed: Returned false for 1001 - likely mishandled even length palindromes or trailing zeros. Fix by comparing all digit pairs correctly.

✓ Correctly identified 1001 as a palindrome.

t2_04edge

Input{"x":100}

Expectedfalse

⏱ Performance - must finish in 2000ms

100 is not a palindrome because trailing zeros break symmetry.

💡 Trailing zeros at the end break palindrome property unless number is zero.

💡 Check if last digit is zero but number is not zero itself.

💡 Return false if x % 10 == 0 and x != 0.

Why it failed: Returned true for 100 - failed to handle trailing zero edge case. Fix by returning false if last digit is zero and number is not zero.

✓ Correctly rejected 100 as non-palindrome due to trailing zeros.

t3_01corner

Input{"x":10}

Expectedfalse

⏱ Performance - must finish in 2000ms

10 is not a palindrome; digits differ when reversed.

💡 Check digit-by-digit comparison rather than just first and last digits.

💡 Avoid greedy approach that only compares first and last digits.

💡 Compare all digit pairs until pointers meet.

Why it failed: Returned true for 10 - greedy approach only compared first and last digits. Fix by checking all digit pairs.

✓ Correctly identified 10 as non-palindrome by full digit comparison.

t3_02corner

Input{"x":123321}

Expectedtrue

⏱ Performance - must finish in 2000ms

123321 is a palindrome with even length, reversed half matches other half.

💡 Try reversing half the number instead of full number to avoid overflow.

💡 Check if reversed half equals the remaining half.

💡 Implement logic to reverse digits until half the number is processed.

Why it failed: Returned false for 123321 - likely confused 0/1 knapsack style full reversal with half reversal. Fix by reversing only half the digits and comparing.

✓ Correctly identified 123321 as palindrome using half reversal.

t3_03corner

Input{"x":2147447412}

Expectedtrue

⏱ Performance - must finish in 2000ms

Large palindrome number within 32-bit range, tests overflow handling.

💡 Avoid reversing full number to prevent integer overflow.

💡 Use half reversal technique to safely compare digits.

💡 Stop reversal when reversed half >= remaining half.

Why it failed: Returned false for 2147447412 - failed to handle overflow in full reversal. Fix by reversing only half the digits.

✓ Correctly handled large palindrome without overflow.

t4_01performance

Input{"_description":"n=100 at constraint boundary - executor generates this input"}

Expectednull

⏱ Performance - must finish in 2000ms

Input size n=100 digits (max length for 32-bit int) must complete in O(log n) time within 2 seconds.

💡 Optimal solution runs in O(log10(n)) time by reversing half the number.

💡 Avoid string conversion or full reversal to improve performance.

💡 Use digit extraction and half reversal to meet time constraints.

Why it failed: TLE on large input - likely used O(n) or O(n^2) approach like full string reversal. Fix by implementing half reversal method with O(log n) complexity.

✓ Algorithm runs within time limits using O(log n) half reversal approach.

Practice

1. You are given a string representing a column title as it appears in an Excel sheet (e.g., "A", "AB", "ZY"). Which algorithmic approach guarantees an optimal and efficient conversion of this title to its corresponding column number?

easy

A. Greedy approach that picks the highest value letter first and sums values

B. Iterative conversion treating the string as a base-26 number and accumulating the result

C. Dynamic programming to store intermediate sums for all prefixes

D. Recursive backtracking to try all possible letter combinations

Solution

1. Step 1: Understand the problem as base-26 conversion

   The Excel column title can be viewed as a number in base 26, where 'A' = 1, 'B' = 2, ..., 'Z' = 26.

2. Step 2: Identify the approach that accumulates values iteratively

   The iterative approach multiplies the current result by 26 and adds the value of the current character, effectively converting base-26 to decimal.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Iterative base-26 conversion matches Excel column numbering [OK]

Hint: Treat the string as a base-26 number [OK]

Common Mistakes:

• Using greedy or DP unnecessarily
• Recursion causes overhead
• Misinterpreting letter values

2. Given the following Python code to find the nth ugly number, what is the output when calling nthUglyNumber(4)?

import heapq

def nthUglyNumber(n):
    heap = [1]
    seen = {1}
    ugly = 1
    for _ in range(n):
        ugly = heapq.heappop(heap)
        for factor in [2, 3, 5]:
            new_ugly = ugly * factor
            if new_ugly not in seen:
                seen.add(new_ugly)
                heapq.heappush(heap, new_ugly)
    return ugly

print(nthUglyNumber(4))

easy

A. 6

B. 5

C. 8

D. 4

Solution

1. Step 1: Trace heap pops and pushes for n=4

   Initial heap: [1]. Iteration 1 pops 1, pushes 2,3,5. Iteration 2 pops 2, pushes 4,6,10. Iteration 3 pops 3, pushes 6 (already seen), 9, 15. Iteration 4 pops 4, pushes 8,12,20.

2. Step 2: Identify the 4th popped ugly number

   The popped numbers in order are 1, 2, 3, 4. The 4th ugly number is 4, so nthUglyNumber(4) returns 4.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   4 is the 4th ugly number generated by the heap [OK]

Hint: Heap pops nth ugly number in ascending order [OK]

Common Mistakes:

• Confusing the iteration count with the returned value
• Forgetting duplicates are skipped
• Miscounting the order of popped elements

3. The following code attempts to count primes less than n using an optimized sieve. Identify the line containing the subtle bug that causes incorrect results or inefficiency.

def countPrimes(n):
    if n <= 2:
        return 0
    is_prime = [True] * (n // 2)
    is_prime[0] = False  # 1 is not prime
    for i in range(3, int(n**0.5) + 1, 2):
        if is_prime[i // 2]:
            for j in range(i, n, 2 * i):  # Bug here
                is_prime[j // 2] = False
    return 1 + sum(is_prime)  # add prime 2

medium

A. Line setting is_prime[0] = false

B. Line returning 1 + sum(is_prime)

C. Line starting the inner loop from j = i instead of i * i

D. Line checking if n <= 2 and returning 0

Solution

1. Step 1: Understand inner loop start

   The inner loop should start marking multiples from i*i, not i, to avoid redundant marking.

2. Step 2: Identify impact of bug

   Starting from i causes marking the prime number i itself as non-prime, corrupting results.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Correct sieve starts marking from i*i to preserve prime i [OK]

Hint: Mark multiples from i*i to avoid marking primes themselves [OK]

Common Mistakes:

• Starting marking from i instead of i*i
• Mislabeling 1 as prime
• Forgetting to add prime 2

4. Consider this buggy recursive reverse integer code snippet:

def reverse(x: int) -> int:
    INT_MAX = 2**31 - 1
    INT_MIN = -2**31
    sign = -1 if x < 0 else 1
    x = abs(x)

    def helper(num, rev):
        if num == 0:
            return rev
        pop = num % 10
        # Bug: Missing overflow check here
        rev = rev * 10 + pop
        return helper(num // 10, rev)

    rev = helper(x, 0)
    return sign * rev if rev != 0 else 0

What is the root cause of incorrect behavior in this code?

medium

A. Line 'rev = rev * 10 + pop' lacks overflow check before multiplication

B. Line with 'sign = -1 if x < 0 else 1' does not handle zero correctly

C. Line 'pop = num % 10' incorrectly extracts digits for negative numbers

D. Line 'return sign * rev if rev != 0 else 0' returns zero incorrectly for valid reversed zero

Solution

1. Step 1: Identify missing overflow check

   The code multiplies rev by 10 and adds pop without checking if rev will overflow 32-bit integer limits.

2. Step 2: Consequence of missing check

   Without overflow check before multiplication, the reversed integer can exceed INT_MAX causing incorrect results or runtime errors.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Overflow check must precede rev update to avoid errors [OK]

Hint: Overflow check must happen before rev update [OK]

Common Mistakes:

• Ignoring overflow before multiplying rev by 10
• Assuming sign handling fixes overflow
• Misunderstanding digit extraction for negatives

5. Suppose you want to count primes less than n but now the input n can be up to 10^12, too large for a standard sieve. Which approach is best suited to handle this scenario efficiently?

hard

A. Use a segmented sieve that processes ranges of size sqrt(n) to reduce memory usage.

B. Use the standard sieve of Eratosthenes with a larger memory allocation.

C. Use brute force primality checks for each number up to n.

D. Use a dynamic programming approach to store prime counts up to n.

Solution

1. Step 1: Understand memory constraints

   Standard sieve requires O(n) space, which is infeasible for n=10^12.

2. Step 2: Identify suitable algorithm

   Segmented sieve divides the range into manageable blocks, using O(sqrt(n)) memory and still marking primes efficiently.

3. Step 3: Compare alternatives

   Brute force is too slow, DP is not applicable, and standard sieve is too large.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Segmented sieve is standard for large n prime counting [OK]

Hint: Segmented sieve handles large n with limited memory [OK]

Common Mistakes:

• Trying standard sieve for huge n
• Using brute force for large ranges
• Misapplying DP to prime counting