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Palindroms ar priekšējo ievietošanu

Izmēģiniet to GfG Practice rakstzīmes-pievienojamās-priekš-palindromam' title=

Dotā virknē s, kas sastāv tikai no mazajiem angļu burtiem, atrodiet minimums rakstzīmju skaits, kam jābūt pievienots uz priekšā no s, lai padarītu to par palindromu.
Piezīme: Palindroms ir virkne, kas lasa vienu un to pašu uz priekšu un atpakaļ.

Piemēri:  

Ievade : s = 'abc'
Izvade : 2
Paskaidrojums : Mēs varam izveidot virs virknes palindromu kā "cbabc", pievienojot "b" un "c" priekšā.



Ievade : s = 'aacecaaaa'
Izvade : 2
Paskaidrojums : Mēs varam izveidot virs virknes palindromu kā "aaaacecaaaa", pievienojot virknes priekšā divus a.

Satura rādītājs

[Naīvā pieeja] Visu prefiksu pārbaude — O(n^2) laiks un O(1) telpa

Ideja ir balstīta uz novērojumu, ka mums ir jāatrod garākais prefikss no dotās virknes, kas ir arī palindroms. Tad minimālās priekšējās rakstzīmes, kas jāpievieno, lai izveidotu doto virknes palindromu, būs atlikušās rakstzīmes.

' title= C++
#include    using namespace std; // function to check if the substring s[i...j] is a palindrome bool isPalindrome(string &s int i int j) {  while (i < j) {    // if characters at the ends are not equal   // it's not a palindrome  if (s[i] != s[j]) {  return false;  }  i++;  j--;  }  return true; } int minChar(string &s) {  int cnt = 0;  int i = s.size() - 1;    // iterate from the end of the string checking for the   // longestpalindrome starting from the beginning  while (i >= 0 && !isPalindrome(s 0 i)) {    i--;  cnt++;  }    return cnt; } int main() {  string s = 'aacecaaaa';  cout << minChar(s);  return 0; } 
C
#include  #include  #include  // function to check if the substring s[i...j] is a palindrome bool isPalindrome(char 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; } int minChar(char s[]) {  int cnt = 0;  int i = strlen(s) - 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; } int main() {    char s[] = 'aacecaaaa';  printf('%d' minChar(s));  return 0; } 
Java
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));  } } 
Python
# 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)) 
C#
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));  } } 
JavaScript
// 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)); 

Izvade
2

[Paredzamā 1. pieeja] KMP algoritma lps masīva izmantošana — O(n) laiks un O(n) telpa

Galvenais novērojums ir tāds, ka virknes garākais palindromiskais prefikss kļūst par tā reversa garāko palindromisko sufiksu.
Ja virkne s = 'aacecaaaa', tās apgrieztā revS = 'aaaacecaa'. Garākais s palindromiskais prefikss ir “aacecaa”.

Lai to efektīvi atrastu, mēs izmantojam LPS masīvu no KMP algoritms . Mēs savienojam sākotnējo virkni ar speciālo rakstzīmi un tās reversu: s + '$' + revS.
Šīs kombinētās virknes LPS masīvs palīdz identificēt garāko s prefiksu, kas atbilst revS sufiksam, kas arī apzīmē s palindromisko prefiksu.

Pēdējā LPS masīva vērtība norāda, cik rakstzīmju jau sākumā veido palindromu. Tādējādi minimālais rakstzīmju skaits, kas jāpievieno, lai padarītu s par palindromu, ir s.length() - lps.back().

C++
#include    #include    #include  using namespace std; vector<int> computeLPSArray(string &pat) {  int n = pat.length();  vector<int> lps(n);  // lps[0] is always 0  lps[0] = 0;  int len = 0;  // loop calculates lps[i] for i = 1 to M-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; } // returns minimum character to be added at // front to make string palindrome int minChar(string &s) {  int n = s.length();  string rev = s;  reverse(rev.begin() rev.end());  // get concatenation of string special character  // and reverse string  s = s + '$' + rev;  // get LPS array of this concatenated string  vector<int> lps = computeLPSArray(s);  // by subtracting last entry of lps vector from  // string length we will get our result  return (n - lps.back()); } int main() {  string s = 'aacecaaaa';  cout << minChar(s);  return 0; } 
Java
import java.util.ArrayList; class GfG {  static int[] computeLPSArray(String pat) {  int n = pat.length();  int[] lps = new int[n];  // lps[0] is always 0  lps[0] = 0;  int len = 0;  // loop calculates lps[i] for i = 1 to n-1  int i = 1;  while (i < n) {  // if the characters match increment len  // and set lps[i]  if (pat.charAt(i) == pat.charAt(len)) {  len++;  lps[i] = len;  i++;  }  // if there is a mismatch  else {  // if len is not zero update len to  // the last known prefix length  if (len != 0) {  len = lps[len - 1];  }  // no prefix matches set lps[i] to 0  else {  lps[i] = 0;  i++;  }  }  }  return lps;  }  // returns minimum character to be added at  // front to make string palindrome  static int minChar(String s) {  int n = s.length();  String rev  = new StringBuilder(s).reverse().toString();  // get concatenation of string special character  // and reverse string  s = s + '$' + 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));  } } 
Python
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 + '$' + rev # get LPS array of this concatenated string lps = computeLPSArray(s) # by subtracting last entry of lps array from # string length we will get our result return n - lps[-1] if __name__ == '__main__': s = 'aacecaaaa' print(minChar(s)) 
C#
using System; class GfG {  static int[] computeLPSArray(string pat) {  int n = pat.Length;  int[] lps = new int[n];  // lps[0] is always 0  lps[0] = 0;  int len = 0;  // loop calculates lps[i] for i = 1 to n-1  int i = 1;  while (i < n) {  // if the characters match increment len  // and set lps[i]  if (pat[i] == pat[len]) {  len++;  lps[i] = len;  i++;  }  // if there is a mismatch  else {  // if len is not zero update len to  // the last known prefix length  if (len != 0) {  len = lps[len - 1];  }  // no prefix matches set lps[i] to 0  else {  lps[i] = 0;  i++;  }  }  }  return lps;  }  // minimum character to be added at  // front to make string palindrome  static int minChar(string s) {  int n = s.Length;  char[] charArray = s.ToCharArray();  Array.Reverse(charArray);  string rev = new string(charArray);  // get concatenation of string special character  // and reverse string  s = s + '$' + 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));  } } 
JavaScript
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)); 

Izvade
2

[Paredzamā 2. pieeja] Menačera algoritma izmantošana

Ideja ir izmantot Manacher's algoritms lai efektīvi atrastu visas palindromiskās apakšvirknes lineārā laikā.
Mēs pārveidojam virkni, ievietojot speciālās rakstzīmes (#), lai vienādi apstrādātu gan pāra, gan nepāra garuma palindromus.
Pēc priekšapstrādes mēs skenējam no sākotnējās virknes beigām un izmantojam palindroma rādiusa masīvu, lai pārbaudītu, vai prefikss s[0...i] ir palindroms. Pirmais šāds rādītājs i dod mums garāko palindromisko prefiksu, un mēs atgriežam n - (i + 1) kā minimālās pievienojamās rakstzīmes.

C++
#include    #include  #include  using namespace std; // manacher's algorithm for finding longest  // palindromic substrings class manacher { public:  // array to store palindrome lengths centered   // at each position  vector<int> p;  // modified string with separators and sentinels  string ms;   manacher(string &s) {  ms = '@';  for (char c : s) {  ms += '#' + string(1 c);  }  ms += '#$';  runManacher();  }  // core Manacher's algorithm  void runManacher() {  int n = ms.size();  p.assign(n 0);  int l = 0 r = 0;  for (int i = 1; i < n - 1; ++i) {  if (i < r)  p[i] = min(r - i p[r + l - i]);  // expand around the current center  while (ms[i + 1 + p[i]] == ms[i - 1 - p[i]])  ++p[i];  // update center if palindrome goes beyond  // current right boundary  if (i + p[i] > r) {  l = i - p[i];  r = i + p[i];  }  }  }  // returns the length of the longest palindrome  // centered at given position  int getLongest(int cen int odd) {  int pos = 2 * cen + 2 + !odd;  return p[pos];  }  // checks whether substring s[l...r] is a palindrome  bool check(int l int r) {  int len = r - l + 1;  int longest = getLongest((l + r) / 2 len % 2);  return len <= longest;  } }; // returns the minimum number of characters to add at the  // front to make the given string a palindrome int minChar(string &s) {  int n = s.size();  manacher m(s);  // scan from the end to find the longest   // palindromic prefix  for (int i = n - 1; i >= 0; --i) {  if (m.check(0 i))  return n - (i + 1);  }  return n - 1; } int main() {  string s = 'aacecaaaa';  cout << minChar(s) << endl;  return 0; } 
Java
class GfG {    // manacher's algorithm for finding longest   // palindromic substrings  static class manacher {  // array to store palindrome lengths centered   // at each position  int[] p;  // modified string with separators and sentinels  String ms;  manacher(String s) {  StringBuilder sb = new StringBuilder('@');  for (char c : s.toCharArray()) {  sb.append('#').append(c);  }  sb.append('#$');  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));  } } 
Python
# 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 += '#$' self.p = [] self.runManacher() # core Manacher's algorithm def runManacher(self): n = len(self.ms) self.p = [0] * n l = r = 0 for i in range(1 n - 1): if i < r: self.p[i] = min(r - i self.p[r + l - i]) # expand around the current center while self.ms[i + 1 + self.p[i]] == self.ms[i - 1 - self.p[i]]: self.p[i] += 1 # update center if palindrome goes beyond  # current right boundary if i + self.p[i] > r: l = i - self.p[i] r = i + self.p[i] # returns the length of the longest palindrome  # centered at given position def getLongest(self cen odd): pos = 2 * cen + 2 + (0 if odd else 1) return self.p[pos] # checks whether substring s[l...r] is a palindrome def check(self l r): length = r - l + 1 longest = self.getLongest((l + r) // 2 length % 2) return length <= longest # returns the minimum number of characters to add at the  # front to make the given string a palindrome def minChar(s): n = len(s) m = manacher(s) # scan from the end to find the longest  # palindromic prefix for i in range(n - 1 -1 -1): if m.check(0 i): return n - (i + 1) return n - 1 if __name__ == '__main__': s = 'aacecaaaa' print(minChar(s)) 
C#
using System; class GfG {    // manacher's algorithm for finding longest   // palindromic substrings  class manacher {  // array to store palindrome lengths centered   // at each position  public int[] p;  // modified string with separators and sentinels  public string ms;  public manacher(string s) {  ms = '@';  foreach (char c in s) {  ms += '#' + c;  }  ms += '#$';  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));  } } 
JavaScript
// 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)); 

Izvade
2 

Laika sarežģītība: O(n) menedžera algoritms darbojas lineārā laikā, palindromus paplašinot katrā centrā bez rakstzīmju atkārtotas apmeklēšanas, un prefiksa pārbaudes cilpa veic O(1) darbības uz katru rakstzīmi virs n rakstzīmēm.
Palīgtelpa: O(n), ko izmanto modificētajai virknei un palindroma garuma masīvam p[], kas abi pieaug lineāri līdz ar ievades lielumu.

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