Dadas dos strings s1 y s2, compruebe si s2 es una rotación de s1.
Ejemplos:
Input : ABACD, CDABA Output : True Input : GEEKS, EKSGE Output : True
Hemos discutido un enfoque en una publicación anterior que maneja la coincidencia de substrings como un patrón. En esta publicación, utilizaremos la construcción lps (prefijo propio más largo que también es sufijo) del algoritmo KMP , que ayudará a encontrar la coincidencia más larga del prefijo de la string b y el sufijo de la string a. Por lo cual conoceremos el punto de giro , a partir de este punto emparejar los personajes. Si todos los caracteres coinciden, entonces es una rotación, de lo contrario no.
A continuación se muestra la implementación básica del enfoque anterior.
C++
// C++ program to check if
// two strings are rotations
// of each other
#include<bits/stdc++.h>
using namespace std;
bool isRotation(string a,
string b)
{
int n = a.length();
int m = b.length();
if (n != m)
return false;
// create lps[] that
// will hold the longest
// prefix suffix values
// for pattern
int lps[n];
// length of the previous
// longest prefix suffix
int len = 0;
int i = 1;
// lps[0] is always 0
lps[0] = 0;
// the loop calculates
// lps[i] for i = 1 to n-1
while (i < n)
{
if (a[i] == b[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 (int k = lps[n - 1];
k < m; ++k)
{
if (b[k] != a[i++])
return false;
}
return true;
}
// Driver code
int main()
{
string s1 = "ABACD";
string s2 = "CDABA";
cout << (isRotation(s1, s2) ?
"1" : "0");
}
// This code is contributed by Chitranayal
Java
// Java program to check if two strings are rotations
// of each other.
import java.util.*;
import java.lang.*;
import java.io.*;
class stringMatching {
public static boolean isRotation(String a, String b)
{
int n = a.length();
int m = b.length();
if (n != m)
return false;
// create lps[] that will hold the longest
// prefix suffix values for pattern
int lps[] = new int[n];
// length of the previous longest prefix suffix
int len = 0;
int 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 (int k = lps[n - 1]; k < m; ++k) {
if (b.charAt(k) != a.charAt(i++))
return false;
}
return true;
}
// Driver code
public static void main(String[] args)
{
String s1 = "ABACD";
String s2 = "CDABA";
System.out.println(isRotation(s1, s2) ? "1" : "0");
}
}
Python3
# Python program to check if
# two strings are rotations
# of each other
def isRotation(a: str, b: str) -> bool:
n = len(a)
m = len(b)
if (n != m):
return False
# create lps[] that
# will hold the longest
# prefix suffix values
# for pattern
lps = [0 for _ in range(n)]
# length of the previous
# longest prefix suffix
length = 0
i = 1
# lps[0] is always 0
lps[0] = 0
# the loop calculates
# lps[i] for i = 1 to n-1
while (i < n):
if (a[i] == b[length]):
length += 1
lps[i] = length
i += 1
else:
if (length == 0):
lps[i] = 0
i += 1
else:
length = lps[length - 1]
i = 0
# Match from that rotating
# point
for k in range(lps[n - 1], m):
if (b[k] != a[i]):
return False
i += 1
return True
# Driver code
if __name__ == "__main__":
s1 = "ABACD"
s2 = "CDABA"
print("1" if isRotation(s1, s2) else "0")
# This code is contributed by sanjeev2552
C#
// C# program to check if
// two strings are rotations
// of each other.
using System;
class GFG
{
public static bool isRotation(string a,
string b)
{
int n = a.Length;
int m = b.Length;
if (n != m)
return false;
// create lps[] that will
// hold the longest prefix
// suffix values for pattern
int []lps = new int[n];
// length of the previous
// longest prefix suffix
int len = 0;
int i = 1;
// lps[0] is always 0
lps[0] = 0;
// the loop calculates
// lps[i] for i = 1 to n-1
while (i < n)
{
if (a[i] == b[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 (int k = lps[n - 1]; k < m; ++k)
{
if (b[k] != a[i++])
return false;
}
return true;
}
// Driver code
public static void Main()
{
string s1 = "ABACD";
string s2 = "CDABA";
Console.WriteLine(isRotation(s1, s2) ?
"1" : "0");
}
}
// This code is contributed
// by anuj_67.
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>
Producción:
1
Complejidad temporal: O(n)
Espacio auxiliar: O(n)