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.

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// 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;
}

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Output:

 4

If you have better and optimized solution for this problem then please share in comments.

This article is contributed by Shashank Mishra ( Gullu ) and reviewed by team geeksforgeeks. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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