Count permutations that produce positive result - GeeksforGeeks

Count permutations that produce positive result - GeeksforGeeks
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

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;

}

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