The naive solution is to first compute all possible permutations of a given string and then iterate over each permutation to see if it’s a palindrome. We can use two pointers to traverse the computed permutation from the beginning and the end simultaneously, comparing each character at every step. If the two pointers point to the same character until they cross each other, the string is a palindrome.
The number of possible permutations for a string of length is , and it requires iterating through the characters to construct each permutation. Therefore, it will take ...
The naive solution is to first compute all possible permutations of a given string and then iterate over each permutation to see if it’s a palindrome. We can use two pointers to traverse the computed permutation from the beginning and the end simultaneously, comparing each character at every step. If the two pointers point to the same character until they cross each other, the string is a palindrome.
The number of possible permutations for a string of length is , and it requires iterating through the characters to construct each permutation. Therefore, it will take ...