LeetCode - valid Palindrome - JAVA - Solution

Welcome to this video on solving the Leetcode Valid Palindrome problem with O(n) time complexity.

In this problem, we are given a string and we need to determine whether it is a valid palindrome or not. A palindrome is a word or phrase that reads the same forwards and backwards.

Instead, we can solve this problem with O(n) time complexity by using two pointers, one at the start of the string and one at the end. We can then iterate through the string, comparing the characters at the two pointers. If they are not the same, we return false. If we reach the end of the string without finding any mismatches, we return true.

Welcome to this video on solving the Leetcode Valid Palindrome problem with O(n) time complexity.

In this problem, we are given a string and we need to determine whether it is a valid palindrome or not. A palindrome is a word or phrase that reads the same forwards and backwards.

One approach to solving this problem would be to loop through the string and check each character to see if it is the same as the corresponding character on the other side of the string. However, this would have a time complexity of O(n^2) as we would need to check each character against all the others.

Instead, we can solve this problem with O(n) time complexity by using two pointers, one at the start of the string and one at the end. We can then iterate through the string, comparing the characters at the two pointers. If they are not the same, we return false. If we reach the end of the string without finding any mismatches, we return true.

This approach has a time complexity of O(n) because we only need to loop through the string once, making it much more efficient than the O(n^2) approach.

I hope this helps give you an understanding of how to solve the Leetcode Valid Palindrome problem with O(n) time complexity. Thanks for watching!