The fastest method for finding the longest palindrome subsequence within a given string

Question

The fastest method for finding the longest palindrome subsequence within a given string

I've been working on creating a function that can identify the longest palindrome subsequence within a given string. After experimenting with dynamic programming, I came up with the code snippet below:

function findLongestPalindrome(str) {
  let length = str.length;
  let memoization = Array(length);
  for (let i = 0; i < length; i++) {
    memoization[i] = Array(length);
    memoization[i][i] = 1;
  }
  for (let cl = 2; cl <= length; cl++) {
    for (let i = 0; i < length - cl + 1; i++) {
      let j = i + cl - 1;
      if (str[i] === str[j] && cl === 2)
        memoization[i][j] = 2;
      else if (str[i] === str[j])
        memoization[i][j] = memoization[i + 1][j - 1] + 2;
      else
        memoization[i][j] = Math.max(memoization[i][j - 1], memoization[i + 1][j]);
    }
  }
  return memoization[0][length - 1];
}

Despite my efforts, this implementation isn't consistently efficient across all test cases and the Time and Space Complexity needs to be optimized further. I've been stuck on this problem for some time now and can't seem to pinpoint the issue. If anyone could provide insight on what might be going wrong and suggest potential solutions, it would be greatly appreciated.

javascript palindrome

Answer 1

Answer №1

According to my analysis, Dynamic Programming may not be the best approach for tackling this particular problem. This is because it does not break down recursively. For instance, when searching for the longest palindrome in a string, you don't necessarily need to consider all second-largest palindromes. Instead, you can simply examine each position and determine if it marks the center of a palindrome longer than any previously encountered. A more efficient way to solve this problem would involve using a greedy algorithm:

const pals = "asd1234321fghjkl1234567887654321qwertzu1234321"

function palindromeAtPos(str, pos, checkEven = false){
  let ix = 0
  const off = checkEven ? 2 : 1
  while(pos-ix-1 >= 0 && pos+ix+1+off < str.length && str[pos-ix-1] === str[pos+ix+off]){
    ix++
  }
  return ix === 0 ? str[pos] : str.substring(pos-ix, pos+ix+off)
}

function longestPalindrome(str){
  let longest = ''
  for(let i = 1; i < str.length; i++){
    const odd = palindromeAtPos(str, i)
    longest = odd.length > longest.length ? odd : longest
    const even = palindromeAtPos(str, i, true)
    longest = even.length > longest.length ? even : longest
  }
  return longest
}

console.log(longestPalindrome(pals))

Although the complexity of this algorithm is quadratic on paper (especially with strings like aaaaaaaaaa), in practice, it tends to exhibit almost linear behavior for most strings.

Answer 2

According to my analysis, Dynamic Programming may not be the best approach for tackling this particular problem. This is because it does not break down recursively. For instance, when searching for the longest palindrome in a string, you don't necessarily need to consider all second-largest palindromes. Instead, you can simply examine each position and determine if it marks the center of a palindrome longer than any previously encountered. A more efficient way to solve this problem would involve using a greedy algorithm:

const pals = "asd1234321fghjkl1234567887654321qwertzu1234321"

function palindromeAtPos(str, pos, checkEven = false){
  let ix = 0
  const off = checkEven ? 2 : 1
  while(pos-ix-1 >= 0 && pos+ix+1+off < str.length && str[pos-ix-1] === str[pos+ix+off]){
    ix++
  }
  return ix === 0 ? str[pos] : str.substring(pos-ix, pos+ix+off)
}

function longestPalindrome(str){
  let longest = ''
  for(let i = 1; i < str.length; i++){
    const odd = palindromeAtPos(str, i)
    longest = odd.length > longest.length ? odd : longest
    const even = palindromeAtPos(str, i, true)
    longest = even.length > longest.length ? even : longest
  }
  return longest
}

console.log(longestPalindrome(pals))

Although the complexity of this algorithm is quadratic on paper (especially with strings like aaaaaaaaaa), in practice, it tends to exhibit almost linear behavior for most strings.

Answer 3

Answer №2

/*
*   x => integer
*   return [] of integers which are multiples of 2
*/
function getMultiplesOfTwo(x) {
   const arr = []
   for (let i = 1; i <= x; i++) {
      if (i % 2 === 0) {
         arr.push(i)
      }
   }
   return arr
}

const num1 = 10
const num2 = 20
const num3 = 30

console.log(getMultiplesOfTwo(num1))
console.log(getMultiplesOfTwo(num2))
console.log(getMultiplesOfTwo(num3))

Answer 4

/*
*   x => integer
*   return [] of integers which are multiples of 2
*/
function getMultiplesOfTwo(x) {
   const arr = []
   for (let i = 1; i <= x; i++) {
      if (i % 2 === 0) {
         arr.push(i)
      }
   }
   return arr
}

const num1 = 10
const num2 = 20
const num3 = 30

console.log(getMultiplesOfTwo(num1))
console.log(getMultiplesOfTwo(num2))
console.log(getMultiplesOfTwo(num3))

The fastest method for finding the longest palindrome subsequence within a given string

Answer №1

Answer №2

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