Count permutations that produce positive result

Given an array of digits of length n > 1, digits lies within range 0 to 9. We perform sequence of below three operations until we are done with all digits

1. Select starting two digits and add ( + )
2. Then next digit is subtracted ( – ) from result of above step. 
   
3. The result of above step is multiplied ( X ) with next digit.

We perform above sequence of operations linearly with remaining digits. 
The task is to find how many permutations of given array that produce positive result after above operations.
For Example, consider input number[] = {1, 2, 3, 4, 5}. Let us consider a permutation 21345 to demonstrate sequence of operations. 

1. Add first two digits, result = 2+1 = 3
2. Subtract next digit, result=result-3= 3-3 = 0
3. Multiply next digit, result=result*4= 0*4 = 0
4. Add next digit, result = result+5 = 0+5 = 5
5. result = 5 which is positive so increment count by one

Examples: 

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 )

Asked in : Morgan Stanley

We first generate all possible permutations of given digit array and perform given sequence of operations sequentially on each permutation and check for which permutation result is positive. Below code describes problem solution easily.
Note : We can generate all possible permutations either by using iterative method, see this article or we can use STL function next_permutation() function to generate it. 

C++

// C++ program to find count of permutations that produce
// positive result.
#include<bits/stdc++.h>
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>

Output:

4

Time Complexity: O(n*n!)

Auxiliary Space: O(1)
If you have better and optimized solution for this problem then please share in comments.