Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome.
Example 1:
Input: "aba" Output: True
Example 2:
Input: "abca" Output: True Explanation: You could delete the character 'c'.
Note:
Intuition and Algorithm
\nFor each index i in the given string, let\'s remove that character, then check if the resulting string is a palindrome. If it is, (or if the original string was a palindrome), then we\'ll return true
Complexity Analysis
\nTime Complexity: where is the length of the string. We do the following times: create a string of length and iterate over it.
\nSpace Complexity: , the space used by our candidate answer.
\nIntuition
\nIf the beginning and end characters of a string are the same (ie. s[0] == s[s.length - 1]), then whether the inner characters are a palindrome (s[1], s[2], ..., s[s.length - 2]) uniquely determines whether the entire string is a palindrome.
Algorithm
\nSuppose we want to know whether s[i], s[i+1], ..., s[j] form a palindrome. If i >= j then we are done. If s[i] == s[j] then we may take i++; j--. Otherwise, the palindrome must be either s[i+1], s[i+2], ..., s[j] or s[i], s[i+1], ..., s[j-1], and we should check both cases.
Complexity Analysis
\nTime Complexity: where is the length of the string. Each of two checks of whether some substring is a palindrome is .
\nSpace Complexity: additional complexity. Only pointers were stored in memory.
\nAnalysis written by: @awice
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