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/*
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Solution:
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1. Use two pointers, one initialised to 0 and the other initialised to end of string. Check if characters at each index
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are the same. If they are the same, shrink both pointers. Else, we have two possibilities: one that neglects character
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at left pointer and the other that neglects character at right pointer. Hence, we check if s[low+1...right] is a palindrome
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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.
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*/
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var validPalindrome = function (s) {
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  let low = 0,
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    high = s.length - 1;
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  while (low < high) {
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    if (s[low] !== s[high]) {
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      return isPalindrome(s, low + 1, high) || isPalindrome(s, low, high - 1);
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    }
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    low++, high--;
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  }
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  return true;
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  // T.C: O(N)
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  // S.C: O(1)
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};
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function isPalindrome(str, low, high) {
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  while (low < high) {
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    if (str[low] !== str[high]) return false;
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    low++, high--;
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  }
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  return true;
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}