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# Largest Palindromic Number - Leetcode Solution

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LeetCode: Largest Palindromic Number Leetcode Solution

Difficulty: Medium

Topics: greedy hash-table string

The Largest Palindromic Number problem on LeetCode is a programming problem, which requires us to find the largest palindromic number that can be obtained by multiplying two n-digit numbers.

A palindromic number is a number that reads the same from both ends. For example, 121, 1221, 12321 are palindromic numbers.

The problem can be solved by iterating over all possible combinations of n-digit numbers and checking if their product is a palindrome. However, this approach is not efficient for large values of n as it takes a lot of time.

A better approach is to start with the largest possible n-digit numbers and systematically decrease one of the numbers until a palindrome is found. Let's take an example to understand this approach better.

Suppose n=3, the largest three-digit number is 999, and the product of 999 and 999 is 998001, which is a palindrome. We can be sure that no other product of two three-digit numbers can produce a palindrome larger than 998001.

Now, let's consider a general case where n=4. The largest four-digit number is 9999, and the product of 9999 and 9999 is 99980001. We can again be sure that no other product of two four-digit numbers can produce a palindrome larger than 99980001.

To implement this approach, we can start with the largest possible n-digit numbers and multiply them together until we find a palindrome. If we don't find a palindrome, we decrease one of the numbers and repeat the process until a palindrome is found or the numbers become equal.

Here is the code implementing this approach:

```
def largest_palindrome(n):
upper_limit = pow(10, n) - 1
lower_limit = pow(10, n-1)
max_palindrome = 0
for i in range(upper_limit, lower_limit-1, -1):
for j in range(i, lower_limit-1, -1):
product = i*j
if product < max_palindrome:
break
if str(product) == str(product)[::-1]:
max_palindrome = product
return max_palindrome
```

In the above code, we start with the largest possible n-digit numbers and decrement one of the numbers until we find a palindrome. If the product of the two numbers is less than the current maximum palindrome, we break out of the loop as there is no need to check further.

This solution has a time complexity of O(n^2) as we are checking all possible combinations of two n-digit numbers. However, this approach is quite efficient for small values of n.

In conclusion, the largest palindromic number problem on LeetCode can be solved using a systematic approach that starts with the largest possible n-digit numbers and decrements one of the numbers until a palindrome is found. This approach has a time complexity of O(n^2) but is quite efficient for small values of n.

## Largest Palindromic Number Solution Code

`1`