As the question suggest. I want an algorithm that runs in linear time which finds the longest prefix of a substring that is a sub-prefix of the same string.


Given a string $S$ of length $n$, we define a function $\delta_{S}:$ $[n]\rightarrow[n]$ as follows. For each integer $w$, $1\le w\le n$, $\delta$ is defined:

$\delta_{S}[w] = \max\{i\colon S[1...i] = S[w...w+i-1]\}$.

The algorithm should find $\delta_{S}[w]$ for each $1\le w \le n$.

Question should be solved using the $\pi$ function of KMP's preprocessing.

The question can be easily solved in $O(n^2)$. The $O(n^2)$ solution is simple and straightforward. For each $w$, manually find the longest prefix. Each iteration could take $O(n)$ time, thus for all $1 \le w \le n$ it would take $O(n^2)$. However, the question should be solved in a linear time. This question falls down under string matching, in particular, KMP. I know the question should be solved using the function $\pi$ from KMP's pre-processing. However, I can't solve the question.

What I tried:

Assume in this answer that all arrays starts from index 1 and ends at index n. Not from 0 until n-1.

Create a new array $\delta$. Initialize $\delta[1]$ = n, since this is always true, and initialize the rest of the array to be 0.

Create an array $\pi$ and compute $\pi(w)$ for $1\le w \le n$, and then for each index, subtract the index of the cell of the array from the content of the array. For instance, $i-\pi(i)$ and then check if this value is greater from $\delta[i]$. However, I seem to get stuck and can't solve the problem. I've tried to solve the problem for a couple of days before posting here, I can't seem to really fully understand the question and how to approach it.

  • $\begingroup$ If you construct a suffix tree/suffix array, does this become easy to solve? $\endgroup$
    – D.W.
    Feb 7, 2023 at 1:27
  • $\begingroup$ Actually yes, it makes easier to solve and I haven't noticed that it becomes much easier to solve it by a suffix tree. However, I'm required to solve this question using the pi function from KMP preprocessing. Additionally, I'd really like to understand how can I really solve it using KMP's preprocessing function. $\endgroup$
    – Mikey
    Feb 7, 2023 at 1:32
  • $\begingroup$ You were asked to compute something called the z-function (cp-algorithms.com/string/z-function.html). $\endgroup$ Feb 7, 2023 at 20:21
  • $\begingroup$ Isn't this the function computed by KMP itself? can you explain how this is different? $\endgroup$ Feb 21, 2023 at 11:20
  • 1
    $\begingroup$ No, it's not the same. KMP pi(i) function computes the longest prefix that is also a suffix of the substring S[0....i]. As stated in the question, given phi(w), we need to find the longest prefix of S that is also a prefix of the substring that starts at S[w]. $\endgroup$
    – Mikey
    Feb 21, 2023 at 14:41

1 Answer 1


The algorithm you mentioned is correct. Note as the comments pointed out, that you want to compute the so called $z$-function: $$ z[i] := \max_{k\in[n]}\{k\,|\, S[1:k] = S[i:i+k],\,\, 0\leq i+k \leq n\} $$ using the $\pi$ function: $$ \pi[i] := \max_{k\in[n]}\{k\,|\, S[1:k] = S[i-k:i],\,\, 0\leq i-k \leq n\} $$ We claim that $$ z[i] := \max_{k\in[n]}\{k\,|\, \pi[i+k] = k,\,\, 0\leq i+k\leq n\} $$ Notice for each $i\in [n]$: $$ \pi[i+k] = k \iff S[1:k] = S[i:i+k] $$ By taking the maximum over $k\in [n]$ our proof is complete. This yields the algorithm:

z[1,...,n] = 0
for i in [n]:
    k = pi(i)
    z[i-k] = max(z[i-k],k)

as you already described.


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