Given two strings s1 and s2, check whether s2 is a rotation of s1.
Examples:
Input : ABACD, CDABA Output : True Input : GEEKS, EKSGE Output : True
We have discussed an approach in earlier post which handles substring match as a pattern. In this post, we will be going to use KMP algorithm’s lps (longest proper prefix which is also suffix) construction, which will help in finding the longest match of the prefix of string b and suffix of string a. By which we will know the rotating point, from this point match the characters. If all the characters are matched, then it is a rotation, else not.
Below is the basic implementation of the above approach.
Javascript
<script> // javascript program to check if two strings are rotations // of each other. function isRotation(a, b) { var n = a.length; var m = b.length; if (n != m) return false; // create lps that will hold the longest // prefix suffix values for pattern var lps = Array.from({length: n}, (_, i) => 0); // length of the previous longest prefix suffix var len = 0; var i = 1; lps[0] = 0; // lps[0] is always 0 // the loop calculates lps[i] for i = 1 to n-1 while (i < n) { if (a.charAt(i) == b.charAt(len)) { lps[i] = ++len; ++i; } else { if (len == 0) { lps[i] = 0; ++i; } else { len = lps[len - 1]; } } } i = 0; // match from that rotating point for (k = lps[n - 1]; k < m; ++k) { if (b.charAt(k) != a.charAt(i++)) return false; } return true; } // Driver code var s1 = "ABACD"; var s2 = "CDABA"; document.write(isRotation(s1, s2) ? "1" : "0"); // This code is contributed by shikhasingrajput. </script> |
Output:
1
Time Complexity: O(n)
Auxiliary Space: O(n)
Please refer complete article on Check if strings are rotations of each other or not | Set 2 for more details!
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