Dada una cadena s que consta únicamente de letras inglesas minúsculas, encuentre la mínimo número de caracteres que deben ser agregado hacia frente de s para convertirlo en un palíndromo.
Nota: Un palíndromo es una cadena que se lee igual hacia adelante y hacia atrás.
Ejemplos:
Aporte : s = 'abc'
Producción : 2
Explicación : Podemos hacer que el palíndromo de cadena anterior sea 'cbabc' agregando 'b' y 'c' al frente.Aporte : s = 'aacecaaaa'
Producción : 2
Explicación : Podemos hacer que el palíndromo de la cuerda anterior sea 'aaaacecaaaa' agregando dos a al frente de la cuerda.
Tabla de contenido
- [Enfoque ingenuo] Comprobación de todos los prefijos: O(n^2) Tiempo y O(1) Espacio
- [Enfoque esperado 1] Uso de la matriz lps del algoritmo KMP: tiempo O(n) y espacio O(n)
- [Enfoque esperado 2] Uso del algoritmo de Manacher
[Enfoque ingenuo] Comprobación de todos los prefijos: O(n^2) Tiempo y O(1) Espacio
La idea se basa en la observación de que necesitamos encontrar el prefijo más largo de una cadena dada que también sea un palíndromo. Luego, los caracteres frontales mínimos que se agregarán para crear un palíndromo de cadena determinado serán los caracteres restantes.
C++ #include
#include
class GfG { // function to check if the substring // s[i...j] is a palindrome static boolean isPalindrome(String s int i int j) { while (i < j) { // if characters at the ends are not the same // it's not a palindrome if (s.charAt(i) != s.charAt(j)) { return false; } i++; j--; } return true; } static int minChar(String s) { int cnt = 0; int i = s.length() - 1; // iterate from the end of the string checking for the // longest palindrome starting from the beginning while (i >= 0 && !isPalindrome(s 0 i)) { i--; cnt++; } return cnt; } public static void main(String[] args) { String s = 'aacecaaaa'; System.out.println(minChar(s)); } }
# function to check if the substring s[i...j] is a palindrome def isPalindrome(s i j): while i < j: # if characters at the ends are not the same # it's not a palindrome if s[i] != s[j]: return False i += 1 j -= 1 return True def minChar(s): cnt = 0 i = len(s) - 1 # iterate from the end of the string checking for the # longest palindrome starting from the beginning while i >= 0 and not isPalindrome(s 0 i): i -= 1 cnt += 1 return cnt if __name__ == '__main__': s = 'aacecaaaa' print(minChar(s))
using System; class GfG { // function to check if the substring s[i...j] is a palindrome static bool isPalindrome(string s int i int j) { while (i < j) { // if characters at the ends are not the same // it's not a palindrome if (s[i] != s[j]) { return false; } i++; j--; } return true; } static int minChar(string s) { int cnt = 0; int i = s.Length - 1; // iterate from the end of the string checking for the longest // palindrome starting from the beginning while (i >= 0 && !isPalindrome(s 0 i)) { i--; cnt++; } return cnt; } static void Main() { string s = 'aacecaaaa'; Console.WriteLine(minChar(s)); } }
// function to check if the substring s[i...j] is a palindrome function isPalindrome(s i j) { while (i < j) { // if characters at the ends are not the same // it's not a palindrome if (s[i] !== s[j]) { return false; } i++; j--; } return true; } function minChar(s) { let cnt = 0; let i = s.length - 1; // iterate from the end of the string checking for the // longest palindrome starting from the beginning while (i >= 0 && !isPalindrome(s 0 i)) { i--; cnt++; } return cnt; } // Driver code let s = 'aacecaaaa'; console.log(minChar(s));
Producción
2
[Enfoque esperado 1] Uso de la matriz lps del algoritmo KMP: tiempo O(n) y espacio O(n)
La observación clave es que el prefijo palindrómico más largo de una cuerda se convierte en el sufijo palindrómico más largo de su reverso.
Dada una cadena s = 'aacecaaaa' su inversa revS = 'aaaacecaa'. El prefijo palindrómico más largo de s es 'aacecaa'.
Para encontrar esto de manera eficiente usamos la matriz LPS del algoritmo KMP . Concatenamos la cadena original con un carácter especial y su reverso: s + '$' + revS.
La matriz LPS para esta cadena combinada ayuda a identificar el prefijo más largo de s que coincide con un sufijo de revS que también representa el prefijo palindrómico de s.
El último valor de la matriz LPS nos dice cuántos caracteres ya forman un palíndromo al principio. Por lo tanto, el número mínimo de caracteres a agregar para hacer de s un palíndromo es s.length() - lps.back().
C++#include
import java.util.ArrayList; class GfG { static int[] computeLPSArray(String pat) { int n = pat.length(); int[] lps = new int[n]; // lps[0] is always 0 lps[0] = 0; int len = 0; // loop calculates lps[i] for i = 1 to n-1 int i = 1; while (i < n) { // if the characters match increment len // and set lps[i] if (pat.charAt(i) == pat.charAt(len)) { len++; lps[i] = len; i++; } // if there is a mismatch else { // if len is not zero update len to // the last known prefix length if (len != 0) { len = lps[len - 1]; } // no prefix matches set lps[i] to 0 else { lps[i] = 0; i++; } } } return lps; } // returns minimum character to be added at // front to make string palindrome static int minChar(String s) { int n = s.length(); String rev = new StringBuilder(s).reverse().toString(); // get concatenation of string special character // and reverse string s = s + '$' + rev; // get LPS array of this concatenated string int[] lps = computeLPSArray(s); // by subtracting last entry of lps array from // string length we will get our result return (n - lps[lps.length - 1]); } public static void main(String[] args) { String s = 'aacecaaaa'; System.out.println(minChar(s)); } }
def computeLPSArray(pat): n = len(pat) lps = [0] * n # lps[0] is always 0 len_lps = 0 # loop calculates lps[i] for i = 1 to n-1 i = 1 while i < n: # if the characters match increment len # and set lps[i] if pat[i] == pat[len_lps]: len_lps += 1 lps[i] = len_lps i += 1 # if there is a mismatch else: # if len is not zero update len to # the last known prefix length if len_lps != 0: len_lps = lps[len_lps - 1] # no prefix matches set lps[i] to 0 else: lps[i] = 0 i += 1 return lps # returns minimum character to be added at # front to make string palindrome def minChar(s): n = len(s) rev = s[::-1] # get concatenation of string special character # and reverse string s = s + '$' + rev # get LPS array of this concatenated string lps = computeLPSArray(s) # by subtracting last entry of lps array from # string length we will get our result return n - lps[-1] if __name__ == '__main__': s = 'aacecaaaa' print(minChar(s))
using System; class GfG { static int[] computeLPSArray(string pat) { int n = pat.Length; int[] lps = new int[n]; // lps[0] is always 0 lps[0] = 0; int len = 0; // loop calculates lps[i] for i = 1 to n-1 int i = 1; while (i < n) { // if the characters match increment len // and set lps[i] if (pat[i] == pat[len]) { len++; lps[i] = len; i++; } // if there is a mismatch else { // if len is not zero update len to // the last known prefix length if (len != 0) { len = lps[len - 1]; } // no prefix matches set lps[i] to 0 else { lps[i] = 0; i++; } } } return lps; } // minimum character to be added at // front to make string palindrome static int minChar(string s) { int n = s.Length; char[] charArray = s.ToCharArray(); Array.Reverse(charArray); string rev = new string(charArray); // get concatenation of string special character // and reverse string s = s + '$' + rev; // get LPS array of this concatenated string int[] lps = computeLPSArray(s); // by subtracting last entry of lps array from // string length we will get our result return n - lps[lps.Length - 1]; } static void Main() { string s = 'aacecaaaa'; Console.WriteLine(minChar(s)); } }
function computeLPSArray(pat) { let n = pat.length; let lps = new Array(n).fill(0); // lps[0] is always 0 let len = 0; // loop calculates lps[i] for i = 1 to n-1 let i = 1; while (i < n) { // if the characters match increment len // and set lps[i] if (pat[i] === pat[len]) { len++; lps[i] = len; i++; } // if there is a mismatch else { // if len is not zero update len to // the last known prefix length if (len !== 0) { len = lps[len - 1]; } // no prefix matches set lps[i] to 0 else { lps[i] = 0; i++; } } } return lps; } // returns minimum character to be added at // front to make string palindrome function minChar(s) { let n = s.length; let rev = s.split('').reverse().join(''); // get concatenation of string special character // and reverse string s = s + '$' + rev; // get LPS array of this concatenated string let lps = computeLPSArray(s); // by subtracting last entry of lps array from // string length we will get our result return n - lps[lps.length - 1]; } // Driver Code let s = 'aacecaaaa'; console.log(minChar(s));
Producción
2
[Enfoque esperado 2] Uso del algoritmo de Manacher
C++La idea es utilizar Algoritmo de Manacher para encontrar eficientemente todas las subcadenas palindrómicas en tiempo lineal.
Transformamos la cadena insertando caracteres especiales (#) para manejar palíndromos de longitud par e impar de manera uniforme.
Después del preprocesamiento, escaneamos desde el final de la cadena original y usamos la matriz de radio palíndromo para verificar si el prefijo s[0...i] es un palíndromo. El primer índice i nos da el prefijo palindrómico más largo y devolvemos n - (i + 1) como los caracteres mínimos a agregar.
#include
class GfG { // manacher's algorithm for finding longest // palindromic substrings static class manacher { // array to store palindrome lengths centered // at each position int[] p; // modified string with separators and sentinels String ms; manacher(String s) { StringBuilder sb = new StringBuilder('@'); for (char c : s.toCharArray()) { sb.append('#').append(c); } sb.append('#$'); ms = sb.toString(); runManacher(); } // core Manacher's algorithm void runManacher() { int n = ms.length(); p = new int[n]; int l = 0 r = 0; for (int i = 1; i < n - 1; ++i) { if (i < r) p[i] = Math.min(r - i p[r + l - i]); // expand around the current center while (ms.charAt(i + 1 + p[i]) == ms.charAt(i - 1 - p[i])) p[i]++; // update center if palindrome goes beyond // current right boundary if (i + p[i] > r) { l = i - p[i]; r = i + p[i]; } } } // returns the length of the longest palindrome // centered at given position int getLongest(int cen int odd) { int pos = 2 * cen + 2 + (odd == 0 ? 1 : 0); return p[pos]; } // checks whether substring s[l...r] is a palindrome boolean check(int l int r) { int len = r - l + 1; int longest = getLongest((l + r) / 2 len % 2); return len <= longest; } } // returns the minimum number of characters to add at the // front to make the given string a palindrome static int minChar(String s) { int n = s.length(); manacher m = new manacher(s); // scan from the end to find the longest // palindromic prefix for (int i = n - 1; i >= 0; --i) { if (m.check(0 i)) return n - (i + 1); } return n - 1; } public static void main(String[] args) { String s = 'aacecaaaa'; System.out.println(minChar(s)); } }
# manacher's algorithm for finding longest # palindromic substrings class manacher: # array to store palindrome lengths centered # at each position def __init__(self s): # modified string with separators and sentinels self.ms = '@' for c in s: self.ms += '#' + c self.ms += '#$' self.p = [] self.runManacher() # core Manacher's algorithm def runManacher(self): n = len(self.ms) self.p = [0] * n l = r = 0 for i in range(1 n - 1): if i < r: self.p[i] = min(r - i self.p[r + l - i]) # expand around the current center while self.ms[i + 1 + self.p[i]] == self.ms[i - 1 - self.p[i]]: self.p[i] += 1 # update center if palindrome goes beyond # current right boundary if i + self.p[i] > r: l = i - self.p[i] r = i + self.p[i] # returns the length of the longest palindrome # centered at given position def getLongest(self cen odd): pos = 2 * cen + 2 + (0 if odd else 1) return self.p[pos] # checks whether substring s[l...r] is a palindrome def check(self l r): length = r - l + 1 longest = self.getLongest((l + r) // 2 length % 2) return length <= longest # returns the minimum number of characters to add at the # front to make the given string a palindrome def minChar(s): n = len(s) m = manacher(s) # scan from the end to find the longest # palindromic prefix for i in range(n - 1 -1 -1): if m.check(0 i): return n - (i + 1) return n - 1 if __name__ == '__main__': s = 'aacecaaaa' print(minChar(s))
using System; class GfG { // manacher's algorithm for finding longest // palindromic substrings class manacher { // array to store palindrome lengths centered // at each position public int[] p; // modified string with separators and sentinels public string ms; public manacher(string s) { ms = '@'; foreach (char c in s) { ms += '#' + c; } ms += '#$'; runManacher(); } // core Manacher's algorithm void runManacher() { int n = ms.Length; p = new int[n]; int l = 0 r = 0; for (int i = 1; i < n - 1; ++i) { if (i < r) p[i] = Math.Min(r - i p[r + l - i]); // expand around the current center while (ms[i + 1 + p[i]] == ms[i - 1 - p[i]]) p[i]++; // update center if palindrome goes beyond // current right boundary if (i + p[i] > r) { l = i - p[i]; r = i + p[i]; } } } // returns the length of the longest palindrome // centered at given position public int getLongest(int cen int odd) { int pos = 2 * cen + 2 + (odd == 0 ? 1 : 0); return p[pos]; } // checks whether substring s[l...r] is a palindrome public bool check(int l int r) { int len = r - l + 1; int longest = getLongest((l + r) / 2 len % 2); return len <= longest; } } // returns the minimum number of characters to add at the // front to make the given string a palindrome static int minChar(string s) { int n = s.Length; manacher m = new manacher(s); // scan from the end to find the longest // palindromic prefix for (int i = n - 1; i >= 0; --i) { if (m.check(0 i)) return n - (i + 1); } return n - 1; } static void Main() { string s = 'aacecaaaa'; Console.WriteLine(minChar(s)); } }
// manacher's algorithm for finding longest // palindromic substrings class manacher { // array to store palindrome lengths centered // at each position constructor(s) { // modified string with separators and sentinels this.ms = '@'; for (let c of s) { this.ms += '#' + c; } this.ms += '#$'; this.p = []; this.runManacher(); } // core Manacher's algorithm runManacher() { const n = this.ms.length; this.p = new Array(n).fill(0); let l = 0 r = 0; for (let i = 1; i < n - 1; ++i) { if (i < r) this.p[i] = Math.min(r - i this.p[r + l - i]); // expand around the current center while (this.ms[i + 1 + this.p[i]] === this.ms[i - 1 - this.p[i]]) this.p[i]++; // update center if palindrome goes beyond // current right boundary if (i + this.p[i] > r) { l = i - this.p[i]; r = i + this.p[i]; } } } // returns the length of the longest palindrome // centered at given position getLongest(cen odd) { const pos = 2 * cen + 2 + (odd === 0 ? 1 : 0); return this.p[pos]; } // checks whether substring s[l...r] is a palindrome check(l r) { const len = r - l + 1; const longest = this.getLongest(Math.floor((l + r) / 2) len % 2); return len <= longest; } } // returns the minimum number of characters to add at the // front to make the given string a palindrome function minChar(s) { const n = s.length; const m = new manacher(s); // scan from the end to find the longest // palindromic prefix for (let i = n - 1; i >= 0; --i) { if (m.check(0 i)) return n - (i + 1); } return n - 1; } // Driver Code const s = 'aacecaaaa'; console.log(minChar(s));
Producción
2
Complejidad del tiempo: El algoritmo de O(n) manacher se ejecuta en tiempo lineal expandiendo palíndromos en cada centro sin volver a visitar los caracteres y el bucle de verificación de prefijo realiza operaciones O(1) por carácter en n caracteres.
Espacio Auxiliar: O(n) utilizado para la cadena modificada y la matriz de longitud del palíndromo p[], los cuales crecen linealmente con el tamaño de entrada.