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kyle-Snippet: Longset Palindromic Substring
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// Write a function that finds the longest palindromic substring in a string. | |
// ex: findLongstPal('affaptbbcracecarf') returns 'racecar' | |
// O(N^2) | |
function findLongestPalOld(input){ | |
var longestPal = ""; | |
var len = input.length; | |
var isPalindrome = function(palindrome){ | |
return palindrome.split('').reverse().join('') == palindrome; | |
} | |
for (var i =0; i < len; i++) { | |
for (var j=len -1; j>=i; j--){ | |
var supposedPal = input.substring(i,j); | |
if (supposedPal.length > 3 && isPalindrome(supposedPal)){ | |
longestPal = longestPal.length > supposedPal.length ? longestPal : supposedPal; | |
} | |
} | |
} | |
return longestPal; | |
} | |
// O(N) | |
// This one is based on some work that smarter people have done. If anything | |
// I learned more about the Big O notation an how to move towards linear time and space | |
// functions. | |
function findLongestPalNew(seq){ | |
var seqLen = seq.length, | |
l = [], | |
center = 0, | |
longestPal = '', | |
palLen = 0; | |
while (center < seqLen){ | |
if (center > palLen && seq[center - palLen - 1] == seq[center]){ | |
palLen += 2; | |
center += 1; | |
continue; | |
} | |
if (palLen > longestPal.length){ | |
longestPal = seq.substr(center - palLen,palLen); | |
} | |
l.push(palLen); | |
var s = l.length - 2; | |
var e = s - palLen; | |
if (s==e) { | |
broke(); | |
continue; | |
} | |
for (var j = s; j >= e; j--) { | |
d = j - e - 1; | |
if (d<0 || j<0) break; | |
if (l[j] == d){ | |
palLen = d; | |
break; | |
} | |
l.push(Math.min(d,l[j])); | |
} | |
function broke(){ | |
palLen = 1; | |
center += 1; | |
} | |
} | |
l.push(palLen); | |
lLen = l.length; | |
st = lLen - 2; | |
en = st - (2 * seqLen + 1 - lLen); | |
for (var k = st; k >= en; k--) { | |
if (k<0) break; | |
d = k - en - 1; | |
var uh = l[k] || false; | |
l.push(Math.min(d, uh)); | |
}; | |
return longestPal; | |
} |
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