Dada una matriz de dígitos de longitud n > 1 dígitos se encuentra dentro del rango de 0 a 9. Realizamos la secuencia de las siguientes tres operaciones hasta que hayamos terminado con todos los dígitos.
- Seleccione los dos dígitos iniciales y agregue (+)
- Luego se resta el siguiente dígito (-) del resultado del paso anterior.
- El resultado del paso anterior se multiplica (X) por el siguiente dígito.
Realizamos la secuencia de operaciones anterior de forma lineal con los dígitos restantes.
La tarea es encontrar cuántas permutaciones de una matriz dada producen resultados positivos después de las operaciones anteriores.
Por ejemplo, considere ingresar el número [] = {1 2 3 4 5}. Consideremos una permutación 21345 para demostrar la secuencia de operaciones.
- Suma el resultado de los primeros dos dígitos = 2+1 = 3
- Resta el siguiente dígito resultado=resultado-3= 3-3 = 0
- Multiplicar el siguiente dígito resultado=resultado*4= 0*4 = 0
- Sumar el resultado del siguiente dígito = resultado+5 = 0+5 = 5
- resultado = 5 que es positivo, por lo que incrementa el recuento en uno
Ejemplos:
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 )
Preguntado en: Morgan Stanley
Primero generamos todas las permutaciones posibles de una matriz de dígitos dada y realizamos una secuencia dada de operaciones secuencialmente en cada permutación y verificamos qué resultado de permutación es positivo. El siguiente código describe la solución del problema fácilmente.
Nota : Podemos generar todas las permutaciones posibles utilizando el método iterativo. este artículo o podemos usar la función STL siguiente_permutación() función para generarlo.
C++
// C++ program to find count of permutations that produce // positive result. #include
// 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 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# 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..
<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>
Producción:
4
Complejidad del tiempo: O (n*n!)
Espacio auxiliar: O(1)
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