Tæl permutationer, der giver positive resultater
Givet en matrix af cifre med længden n > 1 ligger cifre inden for området 0 til 9. Vi udfører en række af under tre operationer, indtil vi er færdige med alle cifre
- Vælg start to cifre og tilføj ( + )
- Derefter trækkes næste ciffer (- ) fra resultatet af ovenstående trin.
- Resultatet af ovenstående trin ganges ( X ) med næste ciffer.
Vi udfører ovenstående sekvens af operationer lineært med resterende cifre.
Opgaven er at finde ud af, hvor mange permutationer af et givet array, der giver positive resultater efter ovenstående operationer.
For eksempel overveje input nummer[] = {1 2 3 4 5}. Lad os overveje en permutation 21345 for at demonstrere rækkefølgen af operationer.
- Tilføj første to cifre resultat = 2+1 = 3
- Træk det næste ciffer fra resultat=resultat-3= 3-3 = 0
- Multiplicer næste ciffer resultat=resultat*4= 0*4 = 0
- Tilføj næste ciffer resultat = resultat+5 = 0+5 = 5
- resultat = 5, hvilket er positivt, så tæl med én
Eksempler:
Input : number[]='123' Output: 4 // here we have all permutations // 123 --> 1+2 -> 3-3 -> 0 // 132 --> 1+3 -> 4-2 -> 2 ( positive ) // 213 --> 2+1 -> 3-3 -> 0 // 231 --> 2+3 -> 5-1 -> 4 ( positive ) // 312 --> 3+1 -> 4-2 -> 2 ( positive ) // 321 --> 3+2 -> 5-1 -> 4 ( positive ) // total 4 permutations are giving positive result Input : number[]='112' Output: 2 // here we have all permutations possible // 112 --> 1+1 -> 2-2 -> 0 // 121 --> 1+2 -> 3-1 -> 2 ( positive ) // 211 --> 2+1 -> 3-1 -> 2 ( positive )
Spurgte i: Morgan Stanley
Vi genererer først alle mulige permutationer af et givet cifferarray og udfører en given sekvens af operationer sekventielt på hver permutation og kontrollerer, hvilket permutationsresultat der er positivt. Nedenstående kode beskriver problemløsning nemt.
Bemærk: Vi kan generere alle mulige permutationer enten ved at bruge iterativ metode se denne artikel, eller vi kan bruge STL-funktionen næste_permutation() funktion til at generere det.
// C++ program to find count of permutations that produce // positive result. #include using namespace std ; // function to find all permutation after executing given // sequence of operations and whose result value is positive // result > 0 ) number[] is array of digits of length of n int countPositivePermutations ( int number [] int n ) { // First sort the array so that we get all permutations // one by one using next_permutation. sort ( number number + n ); // Initialize result (count of permutations with positive // result) int count = 0 ; // Iterate for all permutation possible and do operation // sequentially in each permutation do { // Stores result for current permutation. First we // have to select first two digits and add them int curr_result = number [ 0 ] + number [ 1 ]; // flag that tells what operation we are going to // perform // operation = 0 ---> addition operation ( + ) // operation = 1 ---> subtraction operation ( - ) // operation = 0 ---> multiplication operation ( X ) // first sort the array of digits to generate all // permutation in sorted manner int operation = 1 ; // traverse all digits for ( int i = 2 ; i < n ; i ++ ) { // sequentially perform + - X operation switch ( operation ) { case 0 : curr_result += number [ i ]; break ; case 1 : curr_result -= number [ i ]; break ; case 2 : curr_result *= number [ i ]; break ; } // next operation (decides case of switch) operation = ( operation + 1 ) % 3 ; } // result is positive then increment count by one if ( curr_result > 0 ) count ++ ; // generate next greater permutation until it is // possible } while ( next_permutation ( number number + n )); return count ; } // Driver program to test the case int main () { int number [] = { 1 2 3 }; int n = sizeof ( number ) / sizeof ( number [ 0 ]); cout < < countPositivePermutations ( number n ); return 0 ; }
Java // Java program to find count of permutations // that produce positive result. import java.util.* ; class GFG { // function to find all permutation after // executing given sequence of operations // and whose result value is positive result > 0 ) // number[] is array of digits of length of n static int countPositivePermutations ( int number [] int n ) { // First sort the array so that we get // all permutations one by one using // next_permutation. Arrays . sort ( number ); // Initialize result (count of permutations // with positive result) int count = 0 ; // Iterate for all permutation possible and // do operation sequentially in each permutation do { // Stores result for current permutation. // First we have to select first two digits // and add them int curr_result = number [ 0 ] + number [ 1 ] ; // flag that tells what operation we are going to // perform // operation = 0 ---> addition operation ( + ) // operation = 1 ---> subtraction operation ( - ) // operation = 0 ---> multiplication operation ( X ) // first sort the array of digits to generate all // permutation in sorted manner int operation = 1 ; // traverse all digits for ( int i = 2 ; i < n ; i ++ ) { // sequentially perform + - X operation switch ( operation ) { case 0 : curr_result += number [ i ] ; break ; case 1 : curr_result -= number [ i ] ; break ; case 2 : curr_result *= number [ i ] ; break ; } // next operation (decides case of switch) operation = ( operation + 1 ) % 3 ; } // result is positive then increment count by one if ( curr_result > 0 ) count ++ ; // generate next greater permutation until // it is possible } while ( next_permutation ( number )); return count ; } static boolean next_permutation ( int [] p ) { for ( int a = p . length - 2 ; a >= 0 ; -- a ) if ( p [ a ] < p [ a + 1 ] ) for ( int b = p . length - 1 ;; -- b ) if ( p [ b ] > p [ a ] ) { int t = p [ a ] ; p [ a ] = p [ b ] ; p [ b ] = t ; for ( ++ a b = p . length - 1 ; a < b ; ++ a -- b ) { t = p [ a ] ; p [ a ] = p [ b ] ; p [ b ] = t ; } return true ; } return false ; } // Driver Code public static void main ( String [] args ) { int number [] = { 1 2 3 }; int n = number . length ; System . out . println ( countPositivePermutations ( number n )); } } // This code is contributed by PrinciRaj1992
Python3 # Python3 program to find count of permutations # that produce positive result. # function to find all permutation after # executing given sequence of operations # and whose result value is positive result > 0 ) # number[] is array of digits of length of n def countPositivePermutations ( number n ): # First sort the array so that we get # all permutations one by one using # next_permutation. number . sort () # Initialize result (count of permutations # with positive result) count = 0 ; # Iterate for all permutation possible and # do operation sequentially in each permutation while True : # Stores result for current permutation. # First we have to select first two digits # and add them curr_result = number [ 0 ] + number [ 1 ]; # flag that tells what operation we are going to # perform # operation = 0 ---> addition operation ( + ) # operation = 1 ---> subtraction operation ( - ) # operation = 0 ---> multiplication operation ( X ) # first sort the array of digits to generate all # permutation in sorted manner operation = 1 ; # traverse all digits for i in range ( 2 n ): # sequentially perform + - X operation if operation == 0 : curr_result += number [ i ]; else if operation == 1 : curr_result -= number [ i ]; else if operation == 2 : curr_result *= number [ i ]; # next operation (decides case of switch) operation = ( operation + 1 ) % 3 ; # result is positive then increment count by one if ( curr_result > 0 ): count += 1 # generate next greater permutation until # it is possible if ( not next_permutation ( number )): break return count ; def next_permutation ( p ): for a in range ( len ( p ) - 2 - 1 - 1 ): if ( p [ a ] < p [ a + 1 ]): for b in range ( len ( p ) - 1 - 1000000000 - 1 ): if ( p [ b ] > p [ a ]): t = p [ a ]; p [ a ] = p [ b ]; p [ b ] = t ; a += 1 b = len ( p ) - 1 while ( a < b ): t = p [ a ]; p [ a ] = p [ b ]; p [ b ] = t ; a += 1 b -= 1 return True ; return False ; # Driver Code if __name__ == '__main__' : number = [ 1 2 3 ] n = len ( number ) print ( countPositivePermutations ( number n )); # This code is contributed by rutvik_56.
C# // C# program to find count of permutations // that produce positive result. using System ; class GFG { // function to find all permutation after // executing given sequence of operations // and whose result value is positive result > 0 ) // number[] is array of digits of length of n static int countPositivePermutations ( int [] number int n ) { // First sort the array so that we get // all permutations one by one using // next_permutation. Array . Sort ( number ); // Initialize result (count of permutations // with positive result) int count = 0 ; // Iterate for all permutation possible and // do operation sequentially in each permutation do { // Stores result for current permutation. // First we have to select first two digits // and add them int curr_result = number [ 0 ] + number [ 1 ]; // flag that tells what operation we are going to // perform // operation = 0 ---> addition operation ( + ) // operation = 1 ---> subtraction operation ( - ) // operation = 0 ---> multiplication operation ( X ) // first sort the array of digits to generate all // permutation in sorted manner int operation = 1 ; // traverse all digits for ( int i = 2 ; i < n ; i ++ ) { // sequentially perform + - X operation switch ( operation ) { case 0 : curr_result += number [ i ]; break ; case 1 : curr_result -= number [ i ]; break ; case 2 : curr_result *= number [ i ]; break ; } // next operation (decides case of switch) operation = ( operation + 1 ) % 3 ; } // result is positive then increment count by one if ( curr_result > 0 ) count ++ ; // generate next greater permutation until // it is possible } while ( next_permutation ( number )); return count ; } static bool next_permutation ( int [] p ) { for ( int a = p . Length - 2 ; a >= 0 ; -- a ) if ( p [ a ] < p [ a + 1 ]) for ( int b = p . Length - 1 ;; -- b ) if ( p [ b ] > p [ a ]) { int t = p [ a ]; p [ a ] = p [ b ]; p [ b ] = t ; for ( ++ a b = p . Length - 1 ; a < b ; ++ a -- b ) { t = p [ a ]; p [ a ] = p [ b ]; p [ b ] = t ; } return true ; } return false ; } // Driver Code static public void Main () { int [] number = { 1 2 3 }; int n = number . Length ; Console . Write ( countPositivePermutations ( number n )); } } // This code is contributed by ajit..
JavaScript < script > // Javascript program to find count of permutations // that produce positive result. // function to find all permutation after // executing given sequence of operations // and whose result value is positive result > 0 ) // number[] is array of digits of length of n function countPositivePermutations ( number n ) { // First sort the array so that we get // all permutations one by one using // next_permutation. number . sort ( function ( a b ){ return a - b }); // Initialize result (count of permutations // with positive result) let count = 0 ; // Iterate for all permutation possible and // do operation sequentially in each permutation do { // Stores result for current permutation. // First we have to select first two digits // and add them let curr_result = number [ 0 ] + number [ 1 ]; // flag that tells what operation we are going to // perform // operation = 0 ---> addition operation ( + ) // operation = 1 ---> subtraction operation ( - ) // operation = 0 ---> multiplication operation ( X ) // first sort the array of digits to generate all // permutation in sorted manner let operation = 1 ; // traverse all digits for ( let i = 2 ; i < n ; i ++ ) { // sequentially perform + - X operation switch ( operation ) { case 0 : curr_result += number [ i ]; break ; case 1 : curr_result -= number [ i ]; break ; case 2 : curr_result *= number [ i ]; break ; } // next operation (decides case of switch) operation = ( operation + 1 ) % 3 ; } // result is positive then increment count by one if ( curr_result > 0 ) count ++ ; // generate next greater permutation until // it is possible } while ( next_permutation ( number )); return count ; } function next_permutation ( p ) { for ( let a = p . length - 2 ; a >= 0 ; -- a ) if ( p [ a ] < p [ a + 1 ]) for ( let b = p . length - 1 ;; -- b ) if ( p [ b ] > p [ a ]) { let t = p [ a ]; p [ a ] = p [ b ]; p [ b ] = t ; for ( ++ a b = p . length - 1 ; a < b ; ++ a -- b ) { t = p [ a ]; p [ a ] = p [ b ]; p [ b ] = t ; } return true ; } return false ; } let number = [ 1 2 3 ]; let n = number . length ; document . write ( countPositivePermutations ( number n )); < /script>
Produktion:
4
Tidskompleksitet: O(n*n!)
Hjælpeplads: O(1)
Hvis du har en bedre og optimeret løsning til dette problem, så del venligst i kommentarer.
