Problem Description

You are given an array of integers called queries and a positive integer intLength. Your task is to find specific palindromes based on these inputs.

For each value queries[i], you need to find the queries[i]-th smallest positive palindrome that has exactly intLength digits.

A palindrome is a number that reads the same forwards and backwards. For example, 121 and 1331 are palindromes. Important constraints:

• Palindromes cannot have leading zeros (so 010 is not valid)
• The palindromes must have exactly intLength digits

Your output should be an array answer where:

• answer[i] contains the queries[i]-th smallest palindrome of length intLength
• If the queries[i]-th palindrome doesn't exist (because there aren't that many palindromes of the given length), then answer[i] should be -1

For example, if intLength = 3, the smallest positive palindromes would be: 101, 111, 121, 131, 141, etc. If queries = [1, 2, 5], then:

• The 1st smallest 3-digit palindrome is 101
• The 2nd smallest 3-digit palindrome is 111
• The 5th smallest 3-digit palindrome is 141
• So the answer would be [101, 111, 141]

The solution leverages the fact that for a palindrome, you only need to determine the first half of the digits (rounding up), and the second half mirrors the first half. This significantly reduces the search space.

Intuition

The key insight is that we don't need to generate all palindromes to find the k-th one. Instead, we can directly compute it by understanding the structure of palindromes.

For a palindrome of length intLength, we only need to determine the first half of the digits (including the middle digit if the length is odd). The second half is just a mirror of the first half. This observation drastically simplifies our problem.

Consider palindromes of different lengths:

• Length 3: 101, 111, 121, ... (determined by first 2 digits: 10, 11, 12, ...)
• Length 4: 1001, 1111, 1221, ... (determined by first 2 digits: 10, 11, 12, ...)
• Length 5: 10001, 10101, 10201, ... (determined by first 3 digits: 100, 101, 102, ...)

We notice that:

1. The number of digits we need to determine is (intLength + 1) // 2
2. These determining digits form a sequence starting from 10^(l-1) where l is the number of digits to determine
3. The k-th palindrome corresponds to the k-th number in this sequence

For example, with intLength = 5:

• We need to determine l = (5 + 1) // 2 = 3 digits
• The sequence starts at 100 (which gives palindrome 10001)
• The 2nd palindrome comes from 101 (giving 10101)
• The 3rd palindrome comes from 102 (giving 10201)

To construct the full palindrome from the first half:

• Take the first half digits
• For the second half, reverse the first half
• If intLength is odd, skip the middle digit when reversing (since it's already included)

The range of valid first-half values is from 10^(l-1) to 10^l - 1. If our query asks for a palindrome beyond this range, we return -1.

Learn more about Math patterns.

Solution Approach

Let's walk through the implementation step by step:

1. Calculate the length of the first half:

   l = (intLength + 1) >> 1

   This uses right shift (>> 1) which is equivalent to dividing by 2. For a palindrome of length intLength, we need l digits to determine the entire palindrome.

2. Determine the valid range for the first half:

   start, end = 10 ** (l - 1), 10**l - 1

   • start: The smallest l-digit number (e.g., 100 for 3 digits)
   • end: The largest l-digit number (e.g., 999 for 3 digits)

   These bounds ensure we don't have leading zeros and stay within the valid range.

3. Process each query:

   for q in queries:
       v = start + q - 1

   For the q-th palindrome, we calculate the corresponding first half value v. Since start is the 1st palindrome's first half, we add q - 1 to get the q-th one.

4. Check validity:

   if v > end:
       ans.append(-1)
       continue

   If v exceeds our upper bound, the q-th palindrome doesn't exist for this length.

5. Construct the palindrome:

   s = str(v)
   s += s[::-1][intLength % 2 :]

   • Convert the first half v to string
   • Reverse the string with s[::-1]
   • If intLength is odd (intLength % 2 = 1), skip the first character of the reversed string (the middle digit) using slice [1:]
   • If intLength is even (intLength % 2 = 0), use the entire reversed string with slice [0:]
   • Concatenate to form the complete palindrome

6. Convert and store result:

   ans.append(int(s))

   Convert the palindrome string back to integer and add to results.

Example walkthrough for intLength = 5, queries = [2]:

• l = (5 + 1) >> 1 = 3
• start = 100, end = 999
• For query 2: v = 100 + 2 - 1 = 101
• s = "101"
• Reverse and skip middle: "101"[::-1][1:] = "01"
• Final palindrome: "101" + "01" = "10101"
• Result: 10101

Example Walkthrough

Let's walk through a concrete example with intLength = 4 and queries = [1, 3, 7].

Step 1: Determine the first half length

• Since we need 4-digit palindromes, we calculate: l = (4 + 1) // 2 = 2
• We only need to determine the first 2 digits

Step 2: Find the valid range

• start = 10^(2-1) = 10 (smallest 2-digit number)
• end = 10^2 - 1 = 99 (largest 2-digit number)
• So our first-half values can range from 10 to 99

Step 3: Process each query

Query 1 (find 1st palindrome):

• Calculate first half: v = 10 + 1 - 1 = 10
• Check validity: 10 ≤ 99 ✓
• Construct palindrome:
  • First half: "10"
  • Reverse: "01"
  • Since intLength is even (4 % 2 = 0), use entire reversed string
  • Result: "10" + "01" = "1001"
• Answer: 1001

Query 3 (find 3rd palindrome):

• Calculate first half: v = 10 + 3 - 1 = 12
• Check validity: 12 ≤ 99 ✓
• Construct palindrome:
  • First half: "12"
  • Reverse: "21"
  • Use entire reversed string (even length)
  • Result: "12" + "21" = "1221"
• Answer: 1221

Query 7 (find 7th palindrome):

• Calculate first half: v = 10 + 7 - 1 = 16
• Check validity: 16 ≤ 99 ✓
• Construct palindrome:
  • First half: "16"
  • Reverse: "61"
  • Result: "16" + "61" = "1661"
• Answer: 1661

Final Result: [1001, 1221, 1661]

This demonstrates how we can directly compute the k-th palindrome without generating all preceding ones. The pattern becomes clear: for 4-digit palindromes, we have:

• 1st: 10→1001
• 2nd: 11→1111
• 3rd: 12→1221
• ...and so on

For an odd-length example with intLength = 3 and queries = [2]:

• l = (3 + 1) // 2 = 2
• First half value: v = 10 + 2 - 1 = 11
• Construct: "11" + "11"[::-1][1:] = "11" + "1" = "111"
• The middle digit (second '1') is shared, not duplicated

Solution Implementation

Time and Space Complexity

Time Complexity: O(m * l) where m is the length of queries and l is the length of intLength.

• The algorithm calculates l = (intLength + 1) >> 1, which is O(1).
• Computing start and end using exponentiation 10 ** (l - 1) and 10**l - 1 takes O(l) time each.
• The main loop iterates through each query in queries, which runs m times.
• Inside each iteration:
  • Computing v = start + q - 1 is O(1).
  • Comparison v > end is O(1).
  • Converting v to string takes O(l) time (since v has approximately l digits).
  • String reversal s[::-1] takes O(l) time.
  • String slicing s[::-1][intLength % 2:] takes O(l) time in the worst case.
  • String concatenation takes O(l) time.
  • Converting the final string back to integer takes O(intLength) time.
• Since each iteration performs O(l) operations and we have m iterations, the total time complexity is O(m * l).

Space Complexity: O(m * intLength)

• Variables l, start, and end use O(1) space.
• The ans list stores up to m integers, where each integer can have up to intLength digits, requiring O(m * intLength) space.
• Within each iteration:
  • String s uses O(l) space.
  • The reversed and sliced string uses O(l) space temporarily.
  • The concatenated palindrome string uses O(intLength) space.
• The dominant space usage is the ans list containing all results, giving us O(m * intLength) space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Off-by-One Error in Query Indexing

A frequent mistake is forgetting that queries are 1-indexed while our calculation is 0-indexed. Developers might write:

# WRONG: Missing the -1 adjustment
first_half_value = min_half + query

This would give us the (k+1)-th palindrome instead of the k-th palindrome.

Solution: Always remember to subtract 1 from the query:

first_half_value = min_half + query - 1

2. Incorrect Handling of Odd vs Even Length Palindromes

When constructing the palindrome, it's easy to confuse when to skip the middle character. A common error:

# WRONG: Using wrong condition for slicing
palindrome_str = first_half_str + first_half_str[::-1][1:]  # Always skips first char
# or
palindrome_str = first_half_str + first_half_str[::-1][(intLength + 1) % 2:]  # Wrong formula

Solution: Use intLength % 2 for the slice start:

• For odd lengths (e.g., 5): intLength % 2 = 1, skip the middle digit
• For even lengths (e.g., 4): intLength % 2 = 0, use all digits

palindrome_str = first_half_str + first_half_str[::-1][intLength % 2:]

3. Integer Division vs Bit Shift Confusion

While (intLength + 1) >> 1 and (intLength + 1) // 2 are equivalent for positive integers, using regular division / instead of integer division // causes issues:

# WRONG: Regular division returns float
half_length = (intLength + 1) / 2  # Returns 2.5 for intLength=4
min_half = 10 ** (half_length - 1)  # 10 ** 1.5 gives wrong result

Solution: Use integer division or bit shift:

half_length = (intLength + 1) // 2  # or (intLength + 1) >> 1

4. Boundary Check After Construction

Some might check if the palindrome is valid after constructing it:

# INEFFICIENT: Constructing palindrome before checking validity
first_half_value = min_half + query - 1
palindrome_str = str(first_half_value) + str(first_half_value)[::-1][intLength % 2:]
if len(palindrome_str) > intLength:  # Wrong check
    result.append(-1)

Solution: Check boundaries before construction to avoid unnecessary work:

if first_half_value > max_half:
    result.append(-1)
    continue
# Only construct palindrome if valid

5. Forgetting Edge Cases for Single-Digit Palindromes

When intLength = 1, the logic still works but might be confusing:

• half_length = 1
• min_half = 1, max_half = 9
• The slice [intLength % 2:] becomes [1:] which correctly gives an empty string

However, some might try to handle this as a special case unnecessarily, adding complexity and potential bugs.

Solution: Trust the general formula - it handles all cases including intLength = 1 correctly without special casing.