Étant donné qu'un tableau de chiffres de longueur n > 1 chiffres se situe entre 0 et 9. Nous effectuons une séquence de moins de trois opérations jusqu'à ce que nous ayons terminé avec tous les chiffres.
- Sélectionnez les deux premiers chiffres et ajoutez (+)
- Ensuite, le chiffre suivant est soustrait (-) du résultat de l'étape ci-dessus.
- Le résultat de l'étape ci-dessus est multiplié ( X ) par le chiffre suivant.
Nous effectuons la séquence d'opérations ci-dessus de manière linéaire avec les chiffres restants.
La tâche consiste à trouver combien de permutations d'un tableau donné produisent un résultat positif après les opérations ci-dessus.
Par exemple, considérons l'entrée number[] = {1 2 3 4 5}. Considérons une permutation 21345 pour démontrer la séquence d'opérations.
- Additionnez le résultat des deux premiers chiffres = 2+1 = 3
- Soustraire le chiffre suivant result=result-3= 3-3 = 0
- Multiplier le chiffre suivant résultat=résultat*4= 0*4 = 0
- Ajouter le résultat du chiffre suivant = résultat+5 = 0+5 = 5
- résultat = 5 qui est positif donc incrémenter le compte de un
Exemples :
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 )
Demandé dans : Morgan Stanley
Nous générons d’abord toutes les permutations possibles d’un tableau de chiffres donné et effectuons une séquence donnée d’opérations séquentiellement sur chaque permutation et vérifions pour quel résultat de permutation est positif. Le code ci-dessous décrit facilement la solution au problème.
Note : Nous pouvons générer toutes les permutations possibles soit en utilisant la méthode itérative, voir ce article ou nous pouvons utiliser la fonction STL prochaine_permutation() fonction pour le générer.
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>
Sortir:
4
Complexité temporelle : O(n*n !)
Espace auxiliaire : O(1)
Si vous avez une solution meilleure et optimisée pour ce problème, veuillez la partager dans les commentaires.