Solving Leetcode Interviews in Seconds with AI: Find Palindrome With Fixed Length
Introduction
In this blog post, we will explore how to solve the LeetCode problem "2217" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.
Problem Statement
Given an integer array queries and a positive integer intLength, return an array answer where answer[i] is either the queries[i]th smallest positive palindrome of length intLength or -1 if no such palindrome exists. A palindrome is a number that reads the same backwards and forwards. Palindromes cannot have leading zeros. Example 1: Input: queries = [1,2,3,4,5,90], intLength = 3 Output: [101,111,121,131,141,999] Explanation: The first few palindromes of length 3 are: 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, ... The 90th palindrome of length 3 is 999. Example 2: Input: queries = [2,4,6], intLength = 4 Output: [1111,1331,1551] Explanation: The first six palindromes of length 4 are: 1001, 1111, 1221, 1331, 1441, and 1551. Constraints: 1 <= queries.length <= 5 * 104 1 <= queries[i] <= 109 1 <= intLength <= 15
Explanation
- Palindrome Generation: Generate palindromes based on the given
intLength. We construct the first half (or slightly more than half ifintLengthis odd) of the palindrome and then mirror it to create the full palindrome.- Range Check: Determine the range of possible palindromes of length
intLength. If the requested palindrome index (queries[i]) exceeds the number of possible palindromes, return -1. - Efficient Calculation: Calculate the specific palindrome based on the query index, avoiding explicit generation of all palindromes.
- Range Check: Determine the range of possible palindromes of length
- Runtime Complexity: O(n), where n is the length of the
queriesarray. Storage Complexity: O(n) for the output array.
Code
def kthPalindrome(queries, intLength):
"""
Finds the kth smallest positive palindrome of length intLength for each query.
Args:
queries: A list of integers representing the desired palindrome indices.
intLength: The desired length of the palindromes.
Returns:
A list of integers, where each element is the kth smallest palindrome
or -1 if no such palindrome exists.
"""
results = []
for query in queries:
half_length = (intLength + 1) // 2
start = 10**(half_length - 1)
end = 10**half_length - 1
if query > end - start + 1:
results.append(-1)
continue
num = start + query - 1
num_str = str(num)
if intLength % 2 == 0:
palindrome = num_str + num_str[::-1]
else:
palindrome = num_str + num_str[:-1][::-1]
results.append(int(palindrome))
return results