Find the shortest palindrome

So I encountered this problem during my study: ‘shortest palindrome’. The problem reads:

You are given a string s. You can convert s to a palindrome by adding characters in front of it. Return the shortest palindrome you can find by performing this transformation.

Example 1:

Input: s = "aacecaaa"
Output: "aaacecaaa"
Example 2:

Input: s = "abcd"
Output: "dcbabcd"

Constraints:

0 <= s.length <= 5 * 104
s consists of lowercase English letters only.

I tried to solve this problem with some naive algorithm, and it passed but almost failed on time and space complexity contraints. So I was wondering if there would be a specific algorithm that would solve this problem more efficiently.

The gist of this solution lies in the point that we just need to find the longest palindrome substring from the beginning of the substring. If we find that palindrome substring, we can just add the reverse of the rest of the string to the front of it to get the shortest palindrome.

For example, aacecaaa has the longest palindrome substring aacecaa. Therefore, the leftover is a, which can be appended to the front of the original string, which yields aaacecaaa.

If we don’t use the longest palindrome substring, we can see that the answer will not be the shortest. For example, aacecaaa also has the palindrome substring aa at its beginning. Adding the reverse of the leftover to the beggining of the original string will yield aaacecaacecaaa, which is longer than the shortest palindrome we’ve found above.

Now, how can we possibly find the longest palindrome substring from the beginning of the substring in an efficient way?

Inefficient algorithm: bruteforcing

when:

• n = length of the string to be matched
• m = length of the pattern to be matched

Whenever we are naively doing some pattern matching task, there are inefficient steps that would make the algorithm at most O(n*m).

For example, let’s say that you are looking for every occurence of abc in abceabcii. This means n = 3 and m = 9.

Then, the naive approach would be brute forcing each character with the pattern:

package main

import "fmt"

func brute(s string, pattern string) []int {
	n := len(s)
	m := len(pattern)
	var matchingIndices []int

	for i := range s {
		if i+m > n {
			break
		}
		if s[i:i+m] == pattern {
			matchingIndices = append(matchingIndices, i)
		}
	}

	return matchingIndices
}

func main() {
	cases := []string{
		"abceabciiabc",
	}
	for _, c := range cases {
		fmt.Println(brute(c, "abc"))
	}
}

The time complexity will be O(n*m) because we are iterating over the string n times and each time we are iterating over the pattern m times because s[i:i+m] == pattern will take at most O(m) times to complete, as it is checking string equality.

Anyway, it is easy to spot the inefficiency. Let’s take a look at the algorithm step by step. First, we will try to match abc with abc, bce with abc, cea with abc, and then eab with abc and so on. But what we know for sure is that when you encounter abc, you already know that the next two words of length m will start with b and c, which are not the start of the characters we are looking for. So we can safely skip the b and c, and start fresh at currentIndex+m.

Therefore, what’s important is how many characters you can safely skip. But exactly how many characters can you skip?

Longest Proper Prefix which is Suffix (LPS)

To precisely find how many characters can be safely skipped, we need to find the longest proper prefix which is also a suffix. It does sound weird, but it is what it exactly reads.

• A proper prefix is any prefix of a word that is not the word itself.
• A suffix is any suffix of a word that is also the word itself.

Let’s take the word MOM for example:

• Proper prefixes: '', 'M', 'MO'
• Suffixes: '', 'M', 'OM', 'MOM'

Then, we can see that the LPS is M.

The algorithm to find LPS is pretty simple. First, we will make an array (also called an auxiliary array) for storing the length of LPS in each unique prefix, like so:

Given a string ACABACACD,

1. check A is no LPS at all. record 0
2. check AC has no LPS at all. record 0
3. check ACA has an LPS of length 1 which is A. record 1
4. check ACAB has no LPS at all. record 0
5. check ACABA has an LPS of length 1 which is A. record 1
6. check ACABAC has an LPS of length 2 which is AC. record 2
7. check ACABACA has an LPS of length 3 which is ACA. record 3
8. check ACABACAC has an LPS of length 2 which is AC, record 2
9. check ACABACACD has no LPS at all. record 0

The resulting LPS array should then be:

[0, 0, 1, 0, 1, 2, 3, 2, 0]

The KMP algorithm

KMP is a string matching algorithm that runs in O(n+m) times, where

• n = length of the string to be matched
• m = length of the pattern to be matched

This example runs with the following to elaborate the idea of KMP algorithm:

• text: abcxabcdabxabcdabcdabcy
• pattern: abcdabcy

We realize that three characters (abc) match until we get to text[3] and pattern[3].

The aim is to not go back to the previous indices when we find this mismatch. We only want to go forward.

Then we would need to find the LPS inside the pattern string right before pattern[3]: trivially, we know that abc does not have LPS inside itself because all characters are unique.

This means we can start our next comparison directly from text[3] and pattern[3], which are x and a.

We compare x and a and they’re not a match, so we move to the next character. Forward on, we can see that text[4:10], which is abcdab, has a exact match with pattern[:6]. But it’s not an entire match because text[10] and pattern[6] don’t match, which are x and c.

Then we look for the LPS again in the substring of the pattern that had a match: text[4:10], which is abcdab. Easily we can find that ab is an LPS at pattern[0:2] and pattern[4:6]. Having such an LPS means that the characters before text[10], which is x, must be ab, trivially. But what this also means is that because ab is also a prefix of abcdab, we can start our next search from text[10] (x) and pattern[2] (c).

Why? The search always consists of trivial character-by-character match, and the aim is to skip the redundant searches. abcdab of abcdabcy has an LPS as ab and the text we were matching has ... abcdabx .... This means that by the time we get to x, we already know that we don’t have to go back to the beginning of the pattern to start the search over, because we have already matched ab. Therefore, the next search will start from x and pattern[2], which is c.

Again, comparing c and x. Check if pattern[0:2] has an LPS: no. Then we start character matching from x and pattern[0], all fresh again.

Next up, we see that ... xabcdabcdabcy ... has a match of abcdabc with pattern[:7], but it’s not an entire match, leaving y in the last index in the pattern unmatched with d (text[18]). Again, we see if there is an LPS in abcdabc. This time, the LPS is abc. This means that the last three characters that were checked for match are also the three characters at the beginning of the match, so we can safely say that we already have matched the last three characters (abc) in ... xabcdabcdabc ... with the three first characters of the pattern (abc). So we can start pattern matching from the next character: d (text[18]) and pattern[4], which is also d.

Then finally we find the substring abcdabcy at the end of the text with trivial character-by-character comparison.

Finding the LPS: the efficient way

We know how KMP algorithm works, but we didn’t discuss what an efficient way of obtaining an LPS would be when we looked at how to find an LPS. This is how it’s done:

1. Create int variables j and i to track indices of an lps_array of length len(word)
2. Let j track the prefix, and let i track the suffix of each substring.
3. Let j and i all start from index 0. But because index 0 is always not an LPS, just start i from index 1.
4. Check if word[j] == word[i]. If true, lps_array[i] = j + 1 (which is the length of matching suffix = the index of previously matching prefix + 1) and increment j and i by 1. Otherwise, increment i and reset j to lps_array[j - 1] (which is the length of previously matching prefix). Start this step again.

This Youtube video just kills it, so I’ve watched it multiple times. I’ve set it to start from the LPS part.

an LPS can be constructed with the following code:

func createLpsArray(word string) []int {
    wordLen := len(word)
    lpsArray := make([]int, wordLen)

    i := 1
    j := 0

    for i < wordLen {
        if word[j] == word[i] {
            lpsArray[i] = j + 1
            j++
            i++
        // has a prefix longer than 0 already, so check backwards again
        // when there is a mismatch,
        // we will check the index of previous
        // possible prefix.
        // this is possible because every element in the lps array
        // represents the number of LPS in the substring from 
        // the beginning of the string to the index of the element.
        } else if word[j] != word[i] && j != 0 {
            j = lpsArray[j - 1]
        // this means word[j] != word[i] && j == 0
        // there has been no prefix found so far for the current index i, 
        // so just move onto the next character to find the match
        } else {
            i++
        }
    }
    return lpsArray
}

The complexities, where m is the length of the word, are as follows:

• Time complexity: O(m), because the for loop will run 2m times at its worst and the constant 2 can be removed.
• Space complexity: O(m) because lpsArray is [m]int.

Summing everything up together for KMP substring search

Problem: given a b x a b c a b c a b y (spaces used for convenience), give all starting indices where substring(s) of the text contain the pattern a b c a b y.

Get the LPS array

For the pattern a b c a b y:

1. a and b are not the same, so the array would be

   [0, 0, 0, 0, 0, 0]

2. a and c are not the same, so the array will still be

   [0, 0, 0, 0, 0, 0]

3. a and a are the same. Increment i and j together by one.

   [0, 0, 0, 1, 0, 0]

4. b and b are the same. Because lpsArray[i] = lpsArray[j] + 1, lpsArray[i] = 1 + 1 = 2. Increment i and j together by one.

   [0, 0, 0, 1, 2, 0]

5. c and y are not the same. Then, look backwards (j = lpsArray[j - 1]) to see if there is any other prefixes available. j = lpsArray[j - 1] = 0. Again, word[0] = a and a != y. Mark 0 for the last index:

   [0, 0, 0, 1, 2, 0]

Perform the substring search using the KMP algorithm

what we have so far:

a b x a b c a b c a b y
a b c a b y
[0, 0, 0, 1, 2, 0]

1. index 0: a and a are the same, so proceed together
2. index 1: b and b are the same, so proceed together
3. index 2: x and c are not the same. lpsArray[2 - 1] == 0, so there is no prefix found previously. Start fresh again
4. index 3: a and a are the same, so proceed together
5. index 4: b and b are the same, so proceed together
6. index 5: c and c are the same, so proceed together
7. index 6: a and a are the same, so proceed together
8. index 7: b and b are the same, so proceed together
9. index 7: c and y are not the same. But lpsArray[5 - 1] == 2, so there is a previously found prefix, of which length is 2. This means that we already have a match of length 2 right before c in the text being searched, which we can now safely ignore. Therefore, start the match from lpsArray[2], which is c and text[8] which is also c.
10. index 8+: just proceed and we find that the pattern is the substring at the end of the text

This process can be written in code as follows:

package main

import (
	"fmt"
)


func createLpsArray(word string) []int {
    wordLen := len(word)
    lpsArray := make([]int, wordLen)

    i := 1
    j := 0

    for i < wordLen {
        if word[j] == word[i] {
            lpsArray[i] = j + 1
            j++
            i++
        // has a prefix longer than 0 already, so check backwards again
        // when there is a mismatch,
        // we will check the index of previous
        // possible prefix.
        // this is possible because every element in the lps array
        // represents the number of LPS in the substring from 
        // the beginning of the string to the index of the element.
        } else if word[j] != word[i] && j != 0 {
            j = lpsArray[j - 1]
        // this means word[j] != word[i] && j == 0
        // there has been no prefix found so far for the current index i, so just move onto the next
        // character to find the match
        } else {
            i++
        }
    }
    return lpsArray
}

func kmp(text string, pattern string) []int {
    matchingIndices := make([]int, 0)
    // always false
    if len(pattern) > len(text) {
      return matchingIndices
    }
    
    lpsArray := createLpsArray(pattern)

    fmt.Println(lpsArray)

    i := 0
    j := 0
    for i < len(text) {
    // uncomment to debug
    // fmt.Printf("i: %v, j: %v, text[i]: %v, pattern[j]: %v\n", i, j, string(text[i]), string(pattern[j]))
      if text[i] == pattern[j] {
        i++
        j++
        if j == len(pattern) {
          // store the index of the text where the pattern starts inside the text
          matchingIndices = append(matchingIndices, i - j)
          j = lpsArray[j - 1]
        }
      } else if text[i] != pattern[j] && j != 0 {
        j = lpsArray[j - 1]
      // same as text[i] != pattern[j] && j == 0
      } else {
        i++
      }
    }

    return matchingIndices
}

func main() {
  cases := [][]string{
    {"abxabcabcaby", "abcaby"},
    {"aaaaaaaaaaabbb", "cc"},
    {"aaaaaccccccaccaaaaaccbbb", "cc"},{"aabxabcabczabybxabcabcabxaabxabcabcabybcabcabxabcabcabyabyaby", "abcaby"},
  }

  for _, c := range(cases) {
	  fmt.Println(kmp(c[0], c[1]))
  }
}

Complexities

Given:

• n = length of the string to be matched
• m = length of the pattern to be matched

The time complexity to build the LPS array is O(m) as we’ve seen before, and that to find the substring from the text is O(n). Therefore, adding them up together yields O(m + n), which is the time complexity of the KMP substring search.

The space complexity for LPS array is O(m) because it stores length of an array equivalent to m. The KMP search alone has the space complexity of O(1) because it does not require any variables other than constants. Therefore, the space complexity is O(1 + m) which is O(m).

Back to the original problem: find the shortest palindrome

So let’s revisit the problem again:

You are given a string s. You can convert s to a palindrome by adding characters in front of it. Return the shortest palindrome you can find by performing this transformation.

Example 1:

Input: s = "aacecaaa"
Output: "aaacecaaa"
Example 2:

Input: s = "abcd"
Output: "dcbabcd"

Constraints:

0 <= s.length <= 5 * 104
s consists of lowercase English letters only.

But you revisit the problem description and ponder upon it once more. Do you think you can reword it to something relevant to LPS or KMP search while preserving the same goal?

Yes! If you reword it this way:

Find the longest palindrome substring that starts from index 0.

Actually, we mentioned this already in a very similar term at the beginning of this post:

Find the longest palindrome substring from the beginning of the substring.

Yes. So let’s take the first example for explanation. aacecaaa has aacecaa as the longest palindrome substring starting from index 0, so we just need to reverse the rest of the string, which happens to be a, and prepend it to the string, yielding aaacecaaa.

Now, we don’t have to use KMP algorithm to solve this problem efficiently, but we are going to use the LPS table. KMP algorithm itself isn’t really actually relevant to this problem. Only the LPS table is.

What we really need to do is to build a string like:

s + "#" + reverse(s)

and just run createLpsArray() function which we learned how to create with that string.

Solving aacecaaa

For example, for the string: aacecaaa,

1. do s+some_unique_delimiter+reverse(s). some_unique_delimiter could be anything that is not in the set of chars specified to be available in the problem, because it will otherwise mix up the LPS table obviously. For this example, we are using a #:

   aacecaaa#aaacecaa

2. create an LPS table on that string first:

   a  a  c  e  c  a  a  a  #  a  a  a  c  e  c  a  a
   
   [0, 1, 0, 0, 0, 1, 2, 2, 0, 1, 2, 2, 3, 4, 5, 6, 7]

3. at this step, we can then notice that the aacecaaa has the longest palindrome string of length 7 from index 0. In essence, the LPS table of the crafted string works as viewing the reflection of the original string, which helps to find the longest palindrome in an efficient way.

4. we now know the longest palindrome string is 7 from the index 0. Therefore, the answer has to be:

   reverse(s[7:]) + s

   which is

   a + aacecaaa = aaacecaaa

Solving abcd (just to give another example for perfection)

1. do s+some_unique_delimiter+reverse(s). again:

   abcd#dcba

2. create an LPS table on that string first:

    a  b  c  d  #  d  c  b  a
   
   [0, 0, 0, 0, 0, 0, 0, 0, 1]

3. the longest palindromic string has the length of 1. Therefore, this will be the answer:

   reverse(s[1:]) + s

   which is

   d c b a + b c d = dcbabcd

Solution code

Finally, the solution will look like this:

package main

import "fmt"

func createLpsArray(word string) []int {
    wordLen := len(word)
    lpsArray := make([]int, wordLen)

    i := 1
    j := 0

    for i < wordLen {
        if word[j] == word[i] {
            lpsArray[i] = j + 1
            j++
            i++
        // has a prefix longer than 0 already, so check backwards again
        // when there is a mismatch,
        // we will check the index of previous
        // possible prefix.
        // this is possible because every element in the lps array
        // represents the number of LPS in the substring from 
        // the beginning of the string to the index of the element.
        } else if word[j] != word[i] && j != 0 {
            j = lpsArray[j - 1]
        // this means word[j] != word[i] && j == 0
        // there has been no prefix found so far for the current index i, so just move onto the next
        // character to find the match
        } else {
            i++
        }
    }
    return lpsArray
}

func reverse(s string) string {
    r := []rune(s)
    for i, j := 0, len(r)-1; i < len(r)/2; i, j = i+1, j-1 {
        r[i], r[j] = r[j], r[i]
    }
    return string(r)
}

func shortestPalindrome(s string) string {
  // s + # + reverse(s)
  //  a a c e c a a a # a a a c e c a a
  // [0 1 0 0 0 1 2 2 0 1 2 2 3 4 5 6 7]
  reversed := reverse(s)
  lpsArray := createLpsArray(s + "#" + reversed)
  fmt.Println(lpsArray)
  longestPalindromeLength := lpsArray[len(lpsArray) - 1]

  // a aacecaaa
  return reverse(s[longestPalindromeLength:]) + s
}

func main() {
  cases := []string{
    "aacecaaa", "abcd",
  }
  for _, c := range(cases) {
    fmt.Println(shortestPalindrome(c))
  }
}

Conclusion

So that really sums it up. So far we’ve looked at the mechanics of KMP substring search and its core component: LPS table. Although finding the shortest palindrome is not directly related to KMP substring search algorithm, the LPS table can be borrowed from its underlying idea to solve the problem in an efficient way.

Personally, this one was hard. The algorithm is really just a few lines, but it requires a lot of thought process to get there. But nice job me, I understood how it works. Hope this post helps someone else too.

More substring search algorithms

This post is about KMP algorithm. But there are also other algorithms that can do substring search as efficiently. Check these out too:

• Boyer-Moore string-search
• Rabin-Karp substring search

Credits

• https://binary-baba.medium.com/string-matching-kmp-algorithm-27c182efa387
• https://towardsdatascience.com/pattern-search-with-the-knuth-morris-pratt-kmp-algorithm-8562407dba5b
• https://leetcode.com/problems/shortest-palindrome/solution/
• https://youtu.be/GTJr8OvyEVQ
• https://leetcode.com/problems/shortest-palindrome/discuss/60113/Clean-KMP-solution-with-super-detailed-explanation

Written by Joel Mun. Joel likes Typescript, React, Node.js, GoLang, Python, Wasm and more. He also loves to enlarge the boundaries of his knowledge, mainly by reading books and watching lectures on Youtube. Guitar and piano are necessities at his home.