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This approach uses the two-pointer technique to check if the string is a palindrome after removing at most one character. Start with two pointers at the beginning and end of the string. Move inward while the characters at these pointers are equal. If a mismatch occurs, there are two possibilities: either remove the character at the left pointer or the right pointer. If either results in a palindrome, then the string can be considered a valid palindrome after one deletion.
Time Complexity: O(n), where n is the length of the string.
Space Complexity: O(1) as we use a constant amount of extra space.
1class Solution {
2    private boolean isPalindromeRange(String s, int i, int j) {
3        while (i < j) {
4            if (s.charAt(i) != s.charAt(j)) return false;
5            i++;
6            j--;
7        }
8        return true;
9    }
10    public boolean validPalindrome(String s) {
11        int left = 0;
12        int right = s.length() - 1;
13        while (left < right) {
14            if (s.charAt(left) != s.charAt(right)) {
15                return isPalindromeRange(s, left + 1, right) || isPalindromeRange(s, left, right - 1);
16            }
17            left++;
18            right--;
19        }
20        return true;
21    }
22}The Java solution uses similar logic, checking the string from both ends and verifying sub-ranges for palindromes.
This approach uses recursion to accomplish the same task. When encountering the first differing pair of characters, we make two recursive calls: one ignoring the left character and one ignoring the right character. If either recursive call results in a valid palindrome, the whole string can be considered a valid palindrome after a single character deletion.
Time Complexity: O(n)
Space Complexity: O(1) by avoiding deep recursive stacks through tail optimization.
1    public bool ValidPalindrome(string s) {
        return Helper(s, 0, s.Length - 1, false);
    }
    private bool Helper(string s, int left, int right, bool deleted) {
        while (left < right) {
            if (s[left] != s[right]) {
                if (deleted) return false;
                return Helper(s, left + 1, right, true) || Helper(s, left, right - 1, true);
            }
            left++;
            right--;
        }
        return true;
    }
}This C# version implements a recursive function approach, mirroring solutions in other languages but optimizes tail recursion where applicable.