Find the index of a given permutation in the list of permutations in lexicographic order [duplicate]

Answer 1

Multiply the index of the first digit among the digits in the permutation by (n-1)! and add the index of the remaining permutation.

For example, 2 has index 1, and the index of 13 is 0, so the result is 1 * (3-1)! + 0 = 1 * 2 = 2.

Answer 2

• First element in the permutation gives you a subrange of N!
• Remove first element
• Renumber the remaining elements to remove gaps
• Recurse

Answer 3

In C using recursion it will be something like this

int index(char *abc, int size)  
{   
   if(abc ==0 || *abc==0 || size == 0)  
  {  
    return 0;;  
  }  
  return ((abc[0] - '0') * pow(2,size-1)) + index(abc + 1, size -1);   
}

Answer 4

This solution came into my mind (but I didn't test it yet).

The first digit is from range 1 to N, thus you can derive from the first digit whether your permutation is in which block of size (N-1)!

2*(2!) + X where X = 0..2!-1

Then you can recursively apply this to the next digits (which is one of (N-1)! permutations).

So with an arbitrary N you can do the following algorithm:

X = 0
while string <> ""
  X += ((first digit) - 1) * (N-1)!
  decrease all digits in string by 1 which are > first digit
  remove first digit from string
  N -= 1
return X

In your case:

X = 2
s = "213"
X += (2-1) * 2! => 2
s = "12"
X += (1-1) * 1! => 2
s = "1"
X += (1-1) * 0! => 2

Thus this algorithm is O(N^2).