JavaScript Program to Check if strings are rotations of each other or not

Strings are rotations of each other means two strings that can be rotated to either the right or left direction to obtain the other string. In the right rotation characters of the string shift to their following index and for the zeroth index, it takes the character of the last index assuming the string is in a circle. Left rotation is similar to right rotation but in the opposite direction. We will be given two strings and we have to find whether by rotating the characters of the string we can get another or not.

Input

string1: “abcdef” 
string2: “cdefab”

Output 

Yes

Explanation: We can rotate the first string two times toward the left side and get the second string. String1 after the first rotation will be “bcdefa” and in the next rotation, it will be the same as the second string.

Input

String1: “abcdef”
String2: “bcadef” 

Output 

No

Explanation - The maximum number of rotations that we can rotate a string or array without getting the original string is equal to the length of the given string or the array.

Here, after six rotations we cannot get the string2 from the string1, proving it’s impossible to make both strings equal after the maximum number of rotations.

Naive Approach

In this approach, we will just rotate the given string its length number of times and match with the another given string.

Example

// function to rotate the string in the left direction 
function left_rotate(str){

   // splitting the string and then again joining back 
   str = str.substr(1) + str.substr(0,1);
   return str;
}

// function to check if one string is equal to another after certain rotations
function check(str1, str2){

   // checking the size of both strings 
   if(str1.length != str2.length){
      return false;
   } 
   var k = str1.length
   while(k--){
      if(str1 == str2){
         return true;
      }
      str1 = left_rotate(str1);
   }
   return false;
}

// defining the strings
var str1 = "abcdef" 
var str2 = "cdefab"
console.log("The given strings are " + str1 + " and " + str2);

// calling the function 
if(check(str1,str2)){
   console.log("Yes, we can obtain the second string from the given string by rotating it.");
} else{
   console.log("No, we cannot obtain the second string from the given string by rotating it.");
}

// defining the strings
str1 = "abcdef" 
str2 = "bacdef"
console.log("The given strings are " + str1 + " and " + str2);

// calling the function 
if(check(str1,str2)){
   console.log("Yes, we can obtain the second string from the given string by rotating it.");
} else{
   console.log("No, we cannot obtain the second string from the given string by rotating it.");
}

Output

The given strings are abcdef and cdefab
Yes, we can obtain the second string from the given string by rotating it.
The given strings are abcdef and bacdef
No, we cannot obtain the second string from the given string by rotating it.

Time and Space Complexity

The time complexity of the above code is O(N*N), where N is the size of the given string.

The space complexity of the above code is O(1), as we are not using any space.

KMP Algorithm

In this program we will use the KMP algorithm to find the rotations, let us move to the code.

Example

// function to check if one string is equal to another using KMP
function check(str1, str2){

   // checking the size of both strings 
   if(str1.length != str2.length){
      return false;
   } 
   var len = str1.length;
   
   // create lps that will hold the longest prefix
   var lps = Array.from({length: len}, (_, i) => 0);
   
   // length of the previous longest prefix suffix
   var len_p = 0;
   var i = 1;
   lps[0] = 0; 
   
   // the loop calculates lps[i] for i = 1 to n-1
   while (i < len) {
      if (str1.charAt(i) == str2.charAt(len_p)) {
         lps[i] = ++len_p;
         i++;
      } else {
         if (len_p == 0) {
            lps[i] = 0;
            i++;
         } else {
            len_p = lps[len_p - 1];
         }
      }
   }
   i = 0;
   
   // match from that rotating point
   for(var k = lps[len - 1]; k < len; k++) {
      if (str2.charAt(k) != str1.charAt(i++)){
         return false;
      }
   }
   return true;
}

// defining the strings
var str1 = "abcdef" 
var str2 = "cdefab"
console.log("The given strings are " + str1 + " and " + str2);

// calling the function 
if(check(str1,str2)){
   console.log("Yes, we can obtain the second string from the given string by rotating it.");
} else{
   console.log("No, we cannot obtain the second string from the given string by rotating it.");
}

// defining the strings
str1 = "abcdef" 
str2 = "bacdef"
console.log("The given strings are " + str1 + " and " + str2);

// calling the function 
if(check(str1,str2)){
   console.log("Yes, we can obtain the second string from the given string by rotating it.");
} else{
   console.log("No, we cannot obtain the second string from the given string by rotating it.");
}

Output

The given strings are abcdef and cdefab
Yes, we can obtain the second string from the given string by rotating it.
The given strings are abcdef and bacdef
No, we cannot obtain the second string from the given string by rotating it.

Time and Space Complexity

For the above program, both the time and space complexity are O(N). We have used the extra space to store the values in lps array.

Conclusion

In this tutorial, we have implemented a JavaScript program to check whether we can obtain the one given string from another given string by rotating the characters of the string in either left of right direction. We have used the naive approach which costs us the O(N*N) time and O(1) space complexity. Also, we have implemented KMP Algorithm with O(N) time and space complexity.