The Description of the problem

Given a string s, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together.
Example 1:

Input: s = “abab”
Output: true
Explanation: It is the substring “ab” twice.

Example 2:

Input: s = “aba”
Output: false

The solutions via string perspective

class Solution {
public:bool repeatedSubstringPattern(string s) {int n = s.size();string ss = s + s;ss = ss.substr(1, 2 * n - 2);return ss.find(s) != string::npos;}
};

class Solution:def repeatedSubstringPattern(self, s: str) -> bool:n = len(s)ss = (s + s)[1: 2 * n - 1]return s in ss

The Explanations

This solution checks if the given string s can be constructed by repeating a substring of itself. The intuitive idea behind this solution is that if s can be constructed by repeating a substring of itself, then it must appear at least twice in ss, which is formed by concatenating two copies of s and removing the first and last characters. For example, let’s say s = "abab". Then ss = "abababab". After removing the first and last characters, we get ss = "bababa". Since "ab" appears twice in "bababa", we can conclude that "abab" can be constructed by repreating the substring "ab".

The function return true if s is found in ss using the find() method of the C++ string class. If it’s not found, it returns false.

The solutions via iteration

class Solution {
public:bool repeatedSubstringPattern(string s) {int n = s.size();for (int i = 1; i * 2 <= n; i++) {if (n % i == 0) {bool match = true;for (int j = i; j < n; j++) {if (s[j] != s[j - i]) {match = false;}}if (match) return true;}}return false;}
};

class Solution:def repeatedSubstringPattern(self, s: str) -> bool:n = len(s)for i in range(1, n //2 + 1):if n % i == 0:if all(s[j] == s[j - i] for j in range(i, n)):return Truereturn False

The Explanations

The intuitive idea behind this solution to the “Repeated Substring Pattern” problem is to check all possible substrings of the given string s and see if they can be repeated to form the original string. If any such substring is found, it returns true, otherwise it returns false.

The function first calculates the length of s and store it in the variable n. Then it iterates over all possible substring lengths from 1 to n/2. For each substring length i, it checks if n is divisible by i. If it is, then it’s possible that a substring of length i can be repeated to form the original string.

The function then checks if all substrings of length i are equal. If they are, then it returns true, indicating that the original string can be constructed by repeating a substring of itself. If no such substring is found after checking all possible lengths, the function returns false.

For example, let’s say we have a string "abab". The function will first check it its length 4 is divisible by 1. Since it is not divisible by 1, we move on to checking if its length is divisible by 2. Since 4 is divisible by 2, we check if all substrings of length 2 are equal. In this case, "ab" and "ab" are equal so we return ture.

The solutions with next array in KMP

class Solution {
public:bool repeatedSubstringPattern(string s) {int n = s.size();vector next = {0};int pos = 0;for (int i = 1; i < n; i++) {while (pos != 0 && s[pos] != s[i]) {pos = next[pos - 1];}if (s[pos] == s[i]) pos++;next.push_back(pos);}return next[n-1] && (n % (n - next[n-1]) == 0);}
};