Gegeven een reeks cijfers met een lengte n > 1, liggen de cijfers binnen het bereik van 0 tot 9. We voeren een reeks van onderstaande drie bewerkingen uit totdat we klaar zijn met alle cijfers
- Selecteer de eerste twee cijfers en voeg (+) toe
- Vervolgens wordt het volgende cijfer afgetrokken (-) van het resultaat van bovenstaande stap.
- Het resultaat van bovenstaande stap wordt vermenigvuldigd (X) met het volgende cijfer.
We voeren bovenstaande reeks bewerkingen lineair uit met de resterende cijfers.
De taak is om te bepalen hoeveel permutaties van een gegeven array een positief resultaat opleveren na bovenstaande bewerkingen.
Beschouw bijvoorbeeld invoernummer[] = {1 2 3 4 5}. Laten we een permutatie 21345 bekijken om de volgorde van bewerkingen aan te tonen.
- Voeg de eerste twee cijfers toe: resultaat = 2+1 = 3
- Trek het volgende cijfer af: resultaat=resultaat-3= 3-3 = 0
- Vermenigvuldig het volgende cijfer resultaat=resultaat*4= 0*4 = 0
- Voeg volgend cijfer resultaat toe = resultaat+5 = 0+5 = 5
- resultaat = 5, wat positief is, dus verhoog het aantal met één
Voorbeelden:
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 )
Gevraagd in: Morgan Stanley
We genereren eerst alle mogelijke permutaties van een gegeven cijferarray en voeren een gegeven reeks bewerkingen opeenvolgend uit op elke permutatie en controleren welk permutatieresultaat positief is. Onderstaande code beschrijft de probleemoplossing eenvoudig.
Opmerking : We kunnen alle mogelijke permutaties genereren door de iteratieve methode te gebruiken, zie dit artikel of we kunnen de STL-functie gebruiken volgende_permutatie() functie om het te genereren.
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>
Uitgang:
4
Tijdcomplexiteit: O(n*n!)
Hulpruimte: O(1)
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