# From intuitive-but-slow to really-fast-but-a-little-hard-to-comprehend.

• Solution 1:
Let us start with the very naive solution. It uses 188 ms to solve 100 test cases. The idea is that when we see a character in str that matches the very first character of str, we can start to hoping that str is a built by copies of the substring composed by all characters before the reappearance of the its first character.

``````public class Solution {
public boolean repeatedSubstringPattern(String str) {
int l = str.length();
if(l == 1) {
return false;
}
StringBuilder sb = new StringBuilder();
char first = str.charAt(0);
sb.append(first);
int i = 1;
while(i <= l / 2) {
char c = str.charAt(i++);
if(c == first && isCopies(str, sb.toString())) {
return true;
}else {
sb.append(c);
}
}
return false;
}
private boolean isCopies(String str, String substr) {
if(str.length() % substr.length() != 0) {
return false;
}
for(int i = substr.length(); i < str.length(); i += substr.length()){
if(!str.substring(i).startsWith(substr)){
return false;
}
}
return true;
}
}
``````

Solution 2:
The problem of the first solution is that we do not use the knowledge of failed matching, and the Knuth-Morris-Pratt algorithm is a classic example of how knowledge of failed tries can be use to guide future search.

In fact we only need to compute the pattern table (the lps array, see below) in the Knuth-Morris-Pratt algorithm.

The entry lps[i] is the length of the longest proper prefix that is also a suffix of (s[0], ..., s[i]), or equivalently, length of the longest prefix that is also a proper suffix of (s[0], ..., s[i]). lps[0] is 0, since a single - character string has no proper prefix or proper suffix. Here is a very detailed explanation on the KMP algorithm and how lps is computed dynamically.

After we get lps, we relate the property of the lps table to the property of a string being constructed by joining copies of its substring.

One on hand, if str = (s[0], ..., s[km - 1]) is constructed by joining m copies of its substring substr = (s[0], ..., s[k-1]), and assuming that substr is the finest making blockstr can be boiled down to, meaning str is not constructed by joining copies of any proper substring of substr. Then we must have lps[km - 1] equals (m - 1)k.

On the other hand, assuming that the longest proper prefix of string str that is also a suffix, and the remaining string remainderStr obtained by removing prefix from str satisfies the following 3 properties:

1. remainderStr is a proper substring of str,
2. |str| is divisiable by |remainderStr|,
3. remainderStr is a prefix of prefixStr.

We can show by induction that str is constructed by joining copies of remainderStr.
Here is the code. It solve the 100 test cases in 29ms. A great improvement over the native approach! Remember the statement above, since we are going to use it again.

``````public class Solution {
public boolean repeatedSubstringPattern(String str) {
int l = str.length();
int[] lps = new int[l];
int chasing = 0;
chasing++;
}else {
if(chasing > 0) {
chasing = lps[chasing - 1];
}else {
chasing = 0;
}
}
}
int lp = lps[l - 1];
return (lp > 0 && l % (l - lp) == 0 && str.startsWith(str.substring(lp)));
}
}
``````

Solution 3:
Can the problem be solved efficiently without KMP? The following solution runs even faster (23ms on 100 test cases)

``````public class Solution {
public boolean repeatedSubstringPattern(String str) {
int l = str.length();
for(int i = l / 2; i > 0; i--) {
if(l % i == 0) {
String substr = str.substring(0, i);
int j = i;
while(j < l) {
if(!str.substring(j, j + i).equals(substr)){
break;
}else {
j += i;
}
}
if(j == l) {
return true;
}
}
}
return false;
}
}
``````

Solution 4:
Want clearer code that runs even faster ? Here is it. The idea is stated at the end of the explanation for solution 2. Without really find the longest proper prefix that is also a suffix as in solution 2 and see whether the three properties are matched, we just test each remainderStr, from the longest possible that satisfies condition 1 and 2, that whether the corresponding prefix and suffix match each other. It solve 100 test cases in 16ms. So maybe now, you really want to prove the statement since it lead to such a clean and fast solution? It is not hard to prove by induction.

``````public class Solution {
public boolean repeatedSubstringPattern(String str) {
int l = str.length();
for(int i = (l + 1) / 2; i < l; i++) {
if(l % (l - i) == 0) {
String prefix = str.substring(0, i);
String remainder = str.substring(i);
String suffix = str.substring(l - i);
if(str.startsWith(remainder) && suffix.equals(prefix)){
return true;
}
}
}
return false;
``````

• Why in Solution 4 we do this compare: 'str.startsWith(remainder)' ?

It looks redundant, as it should be always true if suffix and prefix are equal...
I had removed it from my code, and all tests were passed as well.

• This post is deleted!

• @Set You are definitely correct! Thank you for pointing it out. I didn't think very carefully.

• Hi, I had a doubt as to why the third condition is required to check in KMP solution.

Let's say the length of string is N. And, LPS[n-1] = L.

Then, if N>0 and N%(N-L) ==0 should suffice to check if the string is periodic? Right?

• This post is deleted!

• @Set

Could you please show the correctness of solution 4 by induction? I think the key to conceive an linear time algorithm for the problem is how it related to kmp algo. It could be really helpful.
it is quite daunting and actually the statement 3 seems redundant.

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