Valid Palindrome II.md

Given a string s, return true if the s can be palindrome after deleting at most one character from it.

Time complexity: O(n)
Space complexity: O(1)

/**
 * @param {string} s
 * @return {boolean}
 */
 
 /*
1. Use two pointers, one initialised to 0 and the other initialised to end of string. Check if characters at each index
are the same. If they are the same, shrink both pointers. Else, we have two possibilities: one that neglects character
at left pointer and the other that neglects character at right pointer. Hence, we check if s[low+1...right] is a palindrome
or s[low...right-1] is a palindrome. If one of them is a palindrome, we know that we can form a palindrome with one deletion 
and return true. Else, we require more than one deletion, and hence we return false.
*/
var validPalindrome = function(s) {
    let low = 0, high = s.length - 1;
    while(low < high) {
        if(s[low] !== s[high]) {
        // findPalin(s, low + 1, high) is NESSECARRY because otherwise cannot pass the case of 'deeee'
            return findPalin(s, low + 1, high) || findPalin(s, low, high - 1);
        }
        low++;
        high--;
    }
    return true;
};

function findPalin(s, low, high) {
    while(low < high) {
        if(s[low] !== s[high]) {
            return false;
        }
        low++;
        high--;
    }
    return true;
}