I have to create a function which takes a string, and it should return true or false based on whether the input consists of a repeated character sequence. The length of the given string is always greater than 1 and the character sequence must have at least one repetition.
"aa" // true(entirely contains two strings "a")
"aaa" //true(entirely contains three string "a")
"abcabcabc" //true(entirely containas three strings "abc")
"aba" //false(At least there should be two same substrings and nothing more)
"ababa" //false("ab" exists twice but "a" is extra so false)
I have created the below function:
function check(str){
if(!(str.length && str.length - 1)) return false;
let temp = '';
for(let i = 0;i<=str.length/2;i++){
temp += str[i]
//console.log(str.replace(new RegExp(temp,"g"),''))
if(!str.replace(new RegExp(temp,"g"),'')) return true;
}
return false;
}
console.log(check('aa')) //true
console.log(check('aaa')) //true
console.log(check('abcabcabc')) //true
console.log(check('aba')) //false
console.log(check('ababa')) //falseChecking of this is part of the real problem. I can't afford a non-efficient solution like this. First of all, it's looping through half of the string.
The second problem is that it is using replace() in each loop which makes it slow. Is there a better solution regarding performance?
There’s a nifty little theorem about strings like these.
A string consists of the same pattern repeated multiple times if and only if the string is a nontrivial rotation of itself.
Here, a rotation means deleting some number of characters from the front of the string and moving them to the back. For example, the string hello could be rotated to form any of these strings:
hello (the trivial rotation)
elloh
llohe
lohel
ohell
To see why this works, first, assume that a string consists of k repeated copies of a string w. Then deleting the first copy of the repeated pattern (w) from the front of the string and tacking it onto the back will give back the same string. The reverse direction is a bit trickier to prove, but the idea is that if you rotate a string and get back what you started with, you can apply that rotation repeatedly to tile the string with multiple copies of the same pattern (that pattern being the string you needed to move to the end to do the rotation).
Now the question is how to check whether this is the case. For that, there’s another beautiful theorem we can use:
If x and y are strings of the same length, then x is a rotation of y if and only if x is a substring of yy.
As an example, we can see that lohel is a rotation of hello as follows:
hellohello
^^^^^
In our case, we know that every string x will always be a substring of xx (it’ll appear twice, once at each copy of x). So basically we just need to check if our string x is a substring of xx without allowing it to match at the first or halfway character. Here’s a one-liner for that:
function check(str) {
return (str + str).indexOf(str, 1) !== str.length;
}
Assuming indexOf is implemented using a fast string matching algorithm, this will run in time O(n), where n is the length of the input string.
Hope this helps!
You can do it by a capturing group and backreference. Just check it's the repetition of the first captured value.
function check(str) {
return /^(.+)\1+$/.test(str)
}
console.log(check('aa')) //true
console.log(check('aaa')) //true
console.log(check('abcabcabc')) //true
console.log(check('aba')) //false
console.log(check('ababa')) //falseIn the above RegExp:
^ and $ stands for start and end anchors to predict the position.(.+) captures any pattern and captures the value(except \n).\1 is backreference of first captured value and \1+ would check for repetition of captured value.For RegExp debugging use: https://regex101.com/r/pqlAuP/1/debugger
Performance : https://jsperf.com/reegx-and-loop/13
Perhaps the fastest algorithmic approach is building a Z-function in linear time:
The Z-function for this string is an array of length n where the i-th element is equal to the greatest number of characters starting from the position i that coincide with the first characters of s.
In other words, z[i] is the length of the longest common prefix between s and the suffix of s starting at i.
C++ implementation for reference:
vector<int> z_function(string s) {
int n = (int) s.length();
vector<int> z(n);
for (int i = 1, l = 0, r = 0; i < n; ++i) {
if (i <= r)
z[i] = min (r - i + 1, z[i - l]);
while (i + z[i] < n && s[z[i]] == s[i + z[i]])
++z[i];
if (i + z[i] - 1 > r)
l = i, r = i + z[i] - 1;
}
return z;
}
JavaScript implementation
Added optimizations - building a half of z-array and early exit
function z_function(s) {
var n = s.length;
var z = Array(n).fill(0);
var i, l, r;
//for our task we need only a half of z-array
for (i = 1, l = 0, r = 0; i <= n/2; ++i) {
if (i <= r)
z[i] = Math.min(r - i + 1, z[i - l]);
while (i + z[i] < n && s[z[i]] == s[i + z[i]])
++z[i];
//we can check condition and return here
if (z[i] + i === n && n % i === 0) return true;
if (i + z[i] - 1 > r)
l = i, r = i + z[i] - 1;
}
return false;
//return z.some((zi, i) => (i + zi) === n && n % i === 0);
}
console.log(z_function("abacabacabac"));
console.log(z_function("abcab"));Then you need to check indexes i that divide n. If you find such i that i+z[i]=n then the string s can be compressed to the length i and you can return true.
For example, for
string s= 'abacabacabac' with length n=12`
z-array is
(0, 0, 1, 0, 8, 0, 1, 0, 4, 0, 1, 0)
and we can find that for
i=4
i+z[i] = 4 + 8 = 12 = n
and
n % i = 12 % 4 = 0`
so s might be represented as substring of length 4 repeated three times.