Donat a corda s la tasca és trobar el mínim personatges per ser annex (inserció al final) fer un palíndrom de corda.
Exemples:
Entrada : s = 'fet'
Sortida : 2
Explicació: Podem fer palíndrom de corda com 'abede no ' afegint no al final de la corda.
Entrada :s = 'aabb'
Sortida : 2
Explicació: Podem fer palíndrom de corda as'aabb aa ' afegint aa al final de la corda.
Taula de continguts
- Comproveu el palíndrom cada vegada: O(n^2) Temps i O(n) Espai
- Utilitzant l'algoritme Knuth Morris Pratt - O(n) Temps i O(n) Espai
Comproveu el palíndrom cada vegada: O(n^2) Temps i O(n) Espai
C++La solució implica progressivament eliminant caràcters del començament de la corda una per una fins que la corda es converteixi en a palíndrom . La resposta serà el nombre total de caràcters eliminats.
Per exemple, considereu la cadena s = ‘aquí’. Primer comprovem si tota la cadena és un palíndrom que no ho és. A continuació, eliminem el primer caràcter que resulta en el cadena 'pregar'. Tornem a comprovar però encara no és un palíndrom. Aleshores eliminem un altre personatge del principi deixant 'ede'. Aquesta vegada la corda és un palíndrom. Per tant el la sortida és 2 que representa el nombre de caràcters eliminats des del principi per aconseguir un palíndrom.
// C++ code to find minimum number // of appends to make string Palindrome #include
// Java code to find minimum number // of appends to make string Palindrome import java.util.*; class GfG { // Function to check if a given string is a palindrome static boolean isPalindrome(String s) { int left = 0 right = s.length() - 1; while (left < right) { if (s.charAt(left) != s.charAt(right)) return false; left++; right--; } return true; } // Function to find the minimum number of // characters to remove from the beginning static int noOfAppends(String s) { int n = s.length(); // Remove characters from the start until // the string becomes a palindrome for (int i = 0; i < n; i++) { if (isPalindrome(s.substring(i))) { // Return the number of characters removed return i; } } // If no palindrome is found remove // all but one character return n - 1; } public static void main(String[] args) { String s = 'abede'; int result = noOfAppends(s); System.out.println(result); } }
# Python code to find minimum number # of appends to make string Palindrome # Function to check if a given string is a palindrome def is_palindrome(s): left right = 0 len(s) - 1 while left < right: if s[left] != s[right]: return False left += 1 right -= 1 return True # Function to find the minimum number of # characters to remove from the beginning def no_of_appends(s): n = len(s) # Remove characters from the start until # the string becomes a palindrome for i in range(n): if is_palindrome(s[i:]): # Return the number of characters # removed return i # If no palindrome is found remove # all but one character return n - 1 if __name__ == '__main__': s = 'abede' result = no_of_appends(s) print(result)
// C# code to find minimum number // of appends to make string Palindrome using System; class GfG { // Function to check if a given string // is a palindrome static bool IsPalindrome(string s) { int left = 0 right = s.Length - 1; while (left < right) { if (s[left] != s[right]) return false; left++; right--; } return true; } // Function to find the minimum number of // characters to remove from the beginning static int NoOfAppends(string s) { int n = s.Length; // Remove characters from the start until // the string becomes a palindrome for (int i = 0; i < n; i++) { if (IsPalindrome(s.Substring(i))) { // Return the number of characters // removed return i; } } // If no palindrome is found remove all but // one character return n - 1; } static void Main(string[] args) { string s = 'abede'; int result = NoOfAppends(s); Console.WriteLine(result); } }
// JavaScript code to find minimum number // of appends to make string Palindrome // Function to check if a given string is a palindrome function isPalindrome(s) { let left = 0 right = s.length - 1; while (left < right) { if (s[left] !== s[right]) return false; left++; right--; } return true; } // Function to find the minimum number of // characters to remove from the beginning function noOfAppends(s) { let n = s.length; // Remove characters from the start until // the string becomes a palindrome for (let i = 0; i < n; i++) { if (isPalindrome(s.substring(i))) { // Return the number of // characters removed return i; } } // If no palindrome is found remove // all but one character return n - 1; } const s = 'abede'; const result = noOfAppends(s); console.log(result);
Sortida
2
Utilitzant l'algoritme Knuth Morris Pratt - O(n) Temps i O(n) Espai
C++La idea bàsica darrere de l'enfocament és que nosaltres calcular el subcadena més gran des del final i la longitud de la corda menys aquest valor és el mínim nombre d'annexos. La lògica és intuïtiva, no cal afegir-hi palíndrom i només els que no formen el palíndrom. Per trobar aquest palíndrom més gran des del final hem revés la cadena calcula el DFA.
El DFA (autòmat finit determinista) esmentat en el context de la Algoritme de Knuth Morris Pratt és un concepte utilitzat per ajudar a trobar el prefix més llarg d'una cadena que també és un sufix i torneu a invertir la cadena (recuperant així la cadena original) i trobeu l'estat final que representa el nombre de coincidències de la cadena amb la cadena venerada i, per tant, obtenim la subcadena més gran que és un palíndrom des del final.
// CPP program for the given approach // using 2D vector for DFA #include
// Java program for the given approach // using 2D array for DFA import java.util.*; class GfG { // Function to build the DFA and precompute the state static int[][] buildDFA(String s) { int n = s.length(); // Number of possible characters (ASCII range) int c = 256; // Initialize 2D array with zeros int[][] dfa = new int[n][c]; int x = 0; dfa[0][s.charAt(0)] = 1; // Build the DFA for the given string for (int i = 1; i < n; i++) { for (int j = 0; j < c; j++) { dfa[i][j] = dfa[x][j]; } dfa[i][s.charAt(i)] = i + 1; x = dfa[x][s.charAt(i)]; } return dfa; } // Function to find the longest overlap // between the string and its reverse static int longestOverlap(int[][] dfa String query) { int ql = query.length(); int state = 0; // Traverse through the query to // find the longest overlap for (int i = 0; i < ql; i++) { state = dfa[state][query.charAt(i)]; } return state; } // Function to find the minimum // number of characters to append static int minAppends(String s) { // Reverse the string String reversedS = new StringBuilder(s).reverse().toString(); // Build the DFA for the reversed string int[][] dfa = buildDFA(reversedS); // Get the longest overlap with the original string int longestOverlapLength = longestOverlap(dfa s); // Minimum characters to append // to make the string a palindrome return s.length() - longestOverlapLength; } public static void main(String[] args) { String s = 'abede'; System.out.println(minAppends(s)); } }
# Python program for the given approach # using 2D list for DFA # Function to build the DFA and precompute the state def buildDFA(s): n = len(s) # Number of possible characters (ASCII range) c = 256 # Initialize 2D list with zeros dfa = [[0] * c for _ in range(n)] x = 0 dfa[0][ord(s[0])] = 1 # Build the DFA for the given string for i in range(1 n): for j in range(c): dfa[i][j] = dfa[x][j] dfa[i][ord(s[i])] = i + 1 x = dfa[x][ord(s[i])] return dfa # Function to find the longest overlap # between the string and its reverse def longestOverlap(dfa query): ql = len(query) state = 0 # Traverse through the query to # find the longest overlap for i in range(ql): state = dfa[state][ord(query[i])] return state # Function to find the minimum # number of characters to append def minAppends(s): # Reverse the string reversedS = s[::-1] # Build the DFA for the reversed string dfa = buildDFA(reversedS) # Get the longest overlap with the # original string longestOverlapLength = longestOverlap(dfa s) # Minimum characters to append # to make the string a palindrome return len(s) - longestOverlapLength if __name__ == '__main__': s = 'abede' print(minAppends(s))
// C# program for the given approach // using 2D array for DFA using System; class GfG { // Function to build the DFA and precompute the state static int[] buildDFA(string s) { int n = s.Length; // Number of possible characters // (ASCII range) int c = 256; // Initialize 2D array with zeros int[] dfa = new int[n c]; int x = 0; dfa[0 s[0]] = 1; // Build the DFA for the given string for (int i = 1; i < n; i++) { for (int j = 0; j < c; j++) { dfa[i j] = dfa[x j]; } dfa[i s[i]] = i + 1; x = dfa[x s[i]]; } return dfa; } // Function to find the longest overlap // between the string and its reverse static int longestOverlap(int[] dfa string query) { int ql = query.Length; int state = 0; // Traverse through the query to // find the longest overlap for (int i = 0; i < ql; i++) { state = dfa[state query[i]]; } return state; } // Function to find the minimum // number of characters to append static int minAppends(string s) { // Reverse the string using char array char[] reversedArray = s.ToCharArray(); Array.Reverse(reversedArray); string reversedS = new string(reversedArray); // Build the DFA for the reversed string int[] dfa = buildDFA(reversedS); // Get the longest overlap with the original string int longestOverlapLength = longestOverlap(dfa s); // Minimum characters to append // to make the string a palindrome return s.Length - longestOverlapLength; } static void Main() { string s = 'abede'; Console.WriteLine(minAppends(s)); } }
// JavaScript program for the given approach // using 2D array for DFA // Function to build the DFA and precompute the state function buildDFA(s) { let n = s.length; // Number of possible characters // (ASCII range) let c = 256; // Initialize 2D array with zeros let dfa = Array.from({ length: n } () => Array(c).fill(0)); let x = 0; dfa[0][s.charCodeAt(0)] = 1; // Build the DFA for the given string for (let i = 1; i < n; i++) { for (let j = 0; j < c; j++) { dfa[i][j] = dfa[x][j]; } dfa[i][s.charCodeAt(i)] = i + 1; x = dfa[x][s.charCodeAt(i)]; } return dfa; } // Function to find the longest overlap // between the string and its reverse function longestOverlap(dfa query) { let ql = query.length; let state = 0; // Traverse through the query to // find the longest overlap for (let i = 0; i < ql; i++) { state = dfa[state][query.charCodeAt(i)]; } return state; } // Function to find the minimum // number of characters to append function minAppends(s) { // Reverse the string let reversedS = s.split('').reverse().join(''); // Build the DFA for the reversed string let dfa = buildDFA(reversedS); // Get the longest overlap with the original string let longestOverlapLength = longestOverlap(dfa s); // Minimum characters to append // to make the string a palindrome return s.length - longestOverlapLength; } let s = 'abede'; console.log(minAppends(s));
Sortida
2
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- Programació Dinàmica | Set 28 (insercions mínimes per formar un palíndrom)