Número feliç
Prova-ho a GfG Practice 
Un nombre es diu feliç si porta a 1 després d'una seqüència de passos en què cada número de pas se substitueix per la suma de quadrats del seu dígit, és a dir, si comencem amb Happy Number i seguim substituint-lo per xifres suma quadrada arribem a 1.
Exemples:
Input: n = 19
Output: True
19 is Happy Number
1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1
As we reached to 1 19 is a Happy Number.
Input: n = 20
Output: FalseUn nombre no serà un número feliç quan fa un bucle en la seva seqüència, és a dir, toca un número en seqüència que ja s'ha tocat. Així, per comprovar si un número és feliç o no, podem mantenir un conjunt si el mateix nombre es torna a produir, marquem el resultat com a no feliç. Una funció senzilla de l'enfocament anterior es pot escriure com a continuació:
C++Java// method return true if n is Happy Number int numSquareSum ( int n ) { int num = 0 ; while ( n != 0 ) { int digit = n % 10 ; num += digit * digit ; n /= 10 ; } return num ; } int isHappyNumber ( int n ) { set < int > st ; while ( 1 ) { n = numSquareSum ( n ); if ( n == 1 ) return true ; if ( st . find ( n ) != st . end ()) return false ; st . insert ( n ); } }Python// method return true if n is Happy Number public static int numSquareSum ( int n ) { int num = 0 ; while ( n != 0 ) { int digit = n % 10 ; num += digit * digit ; n /= 10 ; } return num ; } static boolean isHappyNumber ( int n ) { HashSet < Integer > st = new HashSet <> (); while ( true ) { n = numSquareSum ( n ); if ( n == 1 ) return true ; if ( st . contains ( n )) return false ; st . add ( n ); } } // This code is contributed by Princi SinghC## method return true if n is Happy Number def numSquareSum ( n ): num = 0 while ( n ): digit = n % 10 num = num + digit * digit n = n // 10 return num def isHappyNumber ( n ): st = set () while ( 1 ): n = numSquareSum ( n ) if ( n == 1 ): return True if n not in st : return False st . insert ( n )JavaScript// Method return true if n is Happy Number static int numSquareSum ( int n ) { int num = 0 ; while ( n != 0 ) { int digit = n % 10 ; num += digit * digit ; n /= 10 ; } return num ; } static int isHappyNumber ( int n ) { HashSet < int > st = new HashSet <> (); while ( 1 ) { n = numSquareSum ( n ); if ( n == 1 ) return true ; if ( st . Contains ( n )) return false ; st . Add ( n ); } } // This code is contributed by 29AjayKumarAnàlisi de complexitat:
Complexitat temporal: O(n*log(n)).
Espai auxiliar: O (n) ja que s'utilitza un conjunt addicional per a l'emmagatzematgePodem resoldre aquest problema sense utilitzar espai addicional i aquesta tècnica també es pot utilitzar en alguns altres problemes similars. Si tractem cada nombre com un node i la substitució per un dígit de suma quadrada com un enllaç, aquest problema és el mateix que trobar un bucle en una llista d'enllaços :
Així, com a solució proposada de l'enllaç anterior, mantindrem dos números lents i ràpids, ambdós inicialitzats a partir d'un nombre donat, lent es substitueix un pas a la vegada i ràpid es substitueix dos passos alhora. Si es troben a l'1, el número donat és Happy Number, en cas contrari no.
C++C// C++ program to check a number is a Happy number or not #includeusing namespace std ; // Utility method to return sum of square of digit of n int numSquareSum ( int n ) { int squareSum = 0 ; while ( n ) { squareSum += ( n % 10 ) * ( n % 10 ); n /= 10 ; } return squareSum ; } // method return true if n is Happy number bool isHappynumber ( int n ) { int slow fast ; // initialize slow and fast by n slow = fast = n ; do { // move slow number by one iteration slow = numSquareSum ( slow ); // move fast number by two iteration fast = numSquareSum ( numSquareSum ( fast )); } while ( slow != fast ); // if both number meet at 1 then return true return ( slow == 1 ); } // Driver code to test above methods int main () { int n = 13 ; if ( isHappynumber ( n )) cout < < n < < ' is a Happy number n ' ; else cout < < n < < ' is not a Happy number n ' ; } // This code is contributed by divyeshrabadiya07 Java// C program to check a number is a Happy number or not #include#include // Utility method to return sum of square of digit of n int numSquareSum ( int n ) { int squareSum = 0 ; while ( n ) { squareSum += ( n % 10 ) * ( n % 10 ); n /= 10 ; } return squareSum ; } // method return true if n is Happy number bool isHappynumber ( int n ) { int slow fast ; // initialize slow and fast by n slow = fast = n ; do { // move slow number by one iteration slow = numSquareSum ( slow ); // move fast number by two iteration fast = numSquareSum ( numSquareSum ( fast )); } while ( slow != fast ); // if both number meet at 1 then return true return ( slow == 1 ); } // Driver code to test above methods int main () { int n = 13 ; if ( isHappynumber ( n )) printf ( '%d is a Happy number n ' n ); else printf ( '%d is not a Happy number n ' n ); } // This code is contributed by Sania Kumari Gupta // (kriSania804) Python// Java program to check a number is a Happy // number or not class GFG { // Utility method to return sum of square of // digit of n static int numSquareSum ( int n ) { int squareSum = 0 ; while ( n != 0 ) { squareSum += ( n % 10 ) * ( n % 10 ); n /= 10 ; } return squareSum ; } // method return true if n is Happy number static boolean isHappynumber ( int n ) { int slow fast ; // initialize slow and fast by n slow = fast = n ; do { // move slow number // by one iteration slow = numSquareSum ( slow ); // move fast number // by two iteration fast = numSquareSum ( numSquareSum ( fast )); } while ( slow != fast ); // if both number meet at 1 // then return true return ( slow == 1 ); } // Driver code to test above methods public static void main ( String [] args ) { int n = 13 ; if ( isHappynumber ( n )) System . out . println ( n + ' is a Happy number' ); else System . out . println ( n + ' is not a Happy number' ); } }C## Python3 program to check if a number is a Happy number or not # Utility method to return the sum of squares of digits of n def num_square_sum ( n ): square_sum = 0 while n : square_sum += ( n % 10 ) ** 2 n //= 10 return square_sum # Method returns True if n is a Happy number def is_happy_number ( n ): # Initialize slow and fast pointers slow = n fast = n while True : # Move slow pointer by one iteration slow = num_square_sum ( slow ) # Move fast pointer by two iterations fast = num_square_sum ( num_square_sum ( fast )) if slow != fast : continue else : break # If both pointers meet at 1 then return True return slow == 1 # Driver Code n = 13 if is_happy_number ( n ): print ( n 'is a Happy number' ) else : print ( n 'is not a Happy number' )JavaScript// C# program to check a number // is a Happy number or not using System ; class GFG { // Utility method to return // sum of square of digit of n static int numSquareSum ( int n ) { int squareSum = 0 ; while ( n != 0 ) { squareSum += ( n % 10 ) * ( n % 10 ); n /= 10 ; } return squareSum ; } // method return true if // n is Happy number static bool isHappynumber ( int n ) { int slow fast ; // initialize slow and // fast by n slow = fast = n ; do { // move slow number // by one iteration slow = numSquareSum ( slow ); // move fast number // by two iteration fast = numSquareSum ( numSquareSum ( fast )); } while ( slow != fast ); // if both number meet at 1 // then return true return ( slow == 1 ); } // Driver code public static void Main () { int n = 13 ; if ( isHappynumber ( n )) Console . WriteLine ( n + ' is a Happy number' ); else Console . WriteLine ( n + ' is not a Happy number' ); } } // This code is contributed by anuj_67.PHP< script > // Javascript program to check a number is a Happy // number or not // Utility method to return sum of square of // digit of n function numSquareSum ( n ) { var squareSum = 0 ; while ( n != 0 ) { squareSum += ( n % 10 ) * ( n % 10 ); n = parseInt ( n / 10 ); } return squareSum ; } // method return true if n is Happy number function isHappynumber ( n ) { var slow fast ; // initialize slow and fast by n slow = fast = n ; do { // move slow number // by one iteration slow = numSquareSum ( slow ); // move fast number // by two iteration fast = numSquareSum ( numSquareSum ( fast )); } while ( slow != fast ); // if both number meet at 1 // then return true return ( slow == 1 ); } // Driver code to test above methods var n = 13 ; if ( isHappynumber ( n )) document . write ( n + ' is a Happy number' ); else document . write ( n + ' is not a Happy number' ); // This code contributed by Princi Singh < /script>Sortida:
13 is a Happy NumberAnàlisi de complexitat:
Complexitat temporal:
Espai auxiliar: O(1).C++
Un altre enfocament per resoldre aquest problema sense espai addicional.
Un número no pot ser un nombre feliç si en qualsevol pas la suma del quadrat de xifres obtinguda és un nombre d'una sola xifra excepte 1 o 7 . Això es deu al fet que l'1 i el 7 són els únics números feliços d'un sol dígit. Amb aquesta informació podem desenvolupar un enfocament tal com es mostra al codi següent:C// C++ program to check if a number is a Happy number or // not. #includeusing namespace std ; // Method - returns true if the input is a happy number else // returns false bool isHappynumber ( int n ) { int sum = n x = n ; // This loop executes till the sum of square of digits // obtained is not a single digit number while ( sum > 9 ) { sum = 0 ; // This loop finds the sum of square of digits while ( x > 0 ) { int d = x % 10 ; sum += d * d ; x /= 10 ; } x = sum ; } if ( sum == 7 || sum == 1 ) return true ; return false ; } int main () { int n = 13 ; if ( isHappynumber ( n )) cout < < n < < ' is a Happy number' ; else cout < < n < < ' is not a Happy number' ; return 0 ; } // This code is contributed by Sania Kumari Gupta Java// C program to check if a number is a Happy number or // not. #include#include // Method - returns true if the input is a happy number else // returns false bool isHappynumber ( int n ) { int sum = n x = n ; // This loop executes till the sum of square of digits // obtained is not a single digit number while ( sum > 9 ) { sum = 0 ; // This loop finds the sum of square of digits while ( x > 0 ) { int d = x % 10 ; sum += d * d ; x /= 10 ; } x = sum ; } if ( sum == 7 || sum == 1 ) return true ; return false ; } int main () { int n = 13 ; if ( isHappynumber ( n )) printf ( '%d is a Happy number' n ); else printf ( '%d is not a Happy number' n ); return 0 ; } // This code is contributed by Sania Kumari Gupta Python// This code is contributed by Vansh Sodhi. // Java program to check if a number is a Happy number or // not. class GFG { // method - returns true if the input is a happy // number else returns false static boolean isHappynumber ( int n ) { int sum = n x = n ; // this loop executes till the sum of square of // digits obtained is not a single digit number while ( sum > 9 ) { sum = 0 ; // this loop finds the sum of square of digits while ( x > 0 ) { int d = x % 10 ; sum += d * d ; x /= 10 ; } x = sum ; } if ( sum == 1 || sum == 7 ) return true ; return false ; } // Driver code public static void main ( String [] args ) { int n = 13 ; if ( isHappynumber ( n )) System . out . println ( n + ' is a Happy number' ); else System . out . println ( n + ' is not a Happy number' ); } }C## Python3 program to check if a number is a Happy number or not. # Method - returns true if the input is # a happy number else returns false def isHappynumber ( n ): Sum x = n n # This loop executes till the sum # of square of digits obtained is # not a single digit number while Sum > 9 : Sum = 0 # This loop finds the sum of # square of digits while x > 0 : d = x % 10 Sum += d * d x = int ( x / 10 ) x = Sum if Sum == 1 or Sum == 7 : return True return False n = 13 if isHappynumber ( n ): print ( n 'is a Happy number' ) else : print ( n 'is not a Happy number' ) # This code is contributed by mukesh07.JavaScript// C# program to check if a number // is a Happy number or not. using System ; class GFG { // Method - returns true if the input is // a happy number else returns false static bool isHappynumber ( int n ) { int sum = n x = n ; // This loop executes till the sum // of square of digits obtained is // not a single digit number while ( sum > 9 ) { sum = 0 ; // This loop finds the sum of // square of digits while ( x > 0 ) { int d = x % 10 ; sum += d * d ; x /= 10 ; } x = sum ; } if ( sum == 1 || sum == 7 ) return true ; return false ; } // Driver code public static void Main ( String [] args ) { int n = 13 ; if ( isHappynumber ( n )) Console . WriteLine ( n + ' is a Happy number' ); else Console . WriteLine ( n + ' is not a Happy number' ); } } // This code is contributed by 29AjayKumar
Sortida13 is a Happy numberAnàlisi de complexitat:
Complexitat temporal: O(n*log(n)).
Espai auxiliar: O(1).Vegeu el vostre article que apareix a la pàgina principal de GeeksforGeeks i ajudeu altres Geeks.
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