[The Knuth-Morris-Pratt (KMP)Algorithm](<https://github.com/Shikha-code36/DS-and-algorithm-Patterns/blob/main/Strings/KMP.py>) -Time Complexity - O(n)

Matching Overview
txt = "AAAAABAAABA"
pat = "AAAA"

We compare first window oftxt withpat
txt = "AAAAABAAABA"
pat = "AAAA"  [Initial position]
We find a match. This is same asNaive String Matching.

In the next step, we compare next window oftxt withpat.
txt = "AAAAABAAABA"
pat =  "AAAA" [Pattern shifted one position]
This is where KMP does optimization over Naive. In this
second window, we only compare fourth A of pattern
with fourth character of current window of text to decide
whether current window matches or not. Since we know
first three characters will anyway match, we skipped
matching first three characters.

Need of Preprocessing?
An important question arises from the above explanation,
how to know how many characters to be skipped. To know this,
we pre-process pattern and prepare an integer array
lps[] that tells us the count of characters to be skipped.

Preprocessing Algorithm:

pat[] = "AAACAAAA"

len = 0, i  = 0.
lps[0] is always 0, we move
to i = 1

len = 0, i  = 1.
Since pat[len] and pat[i] match, do len++,
store it in lps[i] and do i++.
len = 1, lps[1] = 1, i = 2

len = 1, i  = 2.
Since pat[len] and pat[i] match, do len++,
store it in lps[i] and do i++.
len = 2, lps[2] = 2, i = 3

len = 2, i  = 3.
Since pat[len] and pat[i] do not match, and len > 0,
set len = lps[len-1] = lps[1] = 1

len = 1, i  = 3.
Since pat[len] and pat[i] do not match and len > 0,
len = lps[len-1] = lps[0] = 0

len = 0, i  = 3.
Since pat[len] and pat[i] do not match and len = 0,
Set lps[3] = 0 and i = 4.
We know that characters pat
len = 0, i  = 4.
Since pat[len] and pat[i] match, do len++,
store it in lps[i] and do i++.
len = 1, lps[4] = 1, i = 5

len = 1, i  = 5.
Since pat[len] and pat[i] match, do len++,
store it in lps[i] and do i++.
len = 2, lps[5] = 2, i = 6

len = 2, i  = 6.
Since pat[len] and pat[i] match, do len++,
store it in lps[i] and do i++.
len = 3, lps[6] = 3, i = 7

len = 3, i  = 7.
Since pat[len] and pat[i] do not match and len > 0,
set len = lps[len-1] = lps[2] = 2

len = 2, i  = 7.
Since pat[len] and pat[i] match, do len++,
store it in lps[i] and do i++.
len = 3, lps[7] = 3, i = 8

We stop here as we have constructed the whole lps[].