Given a number n, find the number of ways to multiply n elements with an associative operation.
Input : 2 Output : 2 For a and b there are two ways to multiply them. 1. (a * b) 2. (b * a) Input : 3 Output : 12
Explanation(Example 2) :
For a, b and c there are 12 ways to multiply them. 1. ((a * b) * c) 2. (a * (b * c)) 3. ((a * c) * b) 4. (a * (c * b)) 5. ((b * a) * c) 6. (b * (a * c)) 7. ((b * c) * a) 8. (b * (c * a)) 9. ((c * a) * b) 10. (c * (a * b)) 11. ((c * b) * a) 12. (c * (b * a))
Approach : First, we try to find out the recurrence relation. From above examples, we can see h(1) = 1, h(2) = 2, h(3) = 12 . Now, for n elements there will be n – 1 multiplications and n – 1 parentheses. And, (a1, a2, …, an ) can be obtained from (a1, a2, …, a(n – 1)) in exactly one of the two ways :
- Take a multiplication (a1, a2, …, a(n – 1))(which has n – 2 multiplications and n – 2 parentheses) and insert the nth element ‘an’ on either side of either factor in one of the n – 2 multiplications. Thus, for each scheme for n – 1 numbers gives 2 * 2 * (n – 2) = 4 * (n – 2) schemes for n numbers in this way.
- Take a multiplication scheme for (a1, a2, .., a(n-1)) and multiply on left or right by (‘an’). Thus, for each each scheme for n – 1 numbers gives two schemes for n numbers in this way.
So after adding above two, we get, h(n) = (4 * n – 8 + 2) * h(n – 1), h(n) = (4 * n – 6) * h(n – 1). This recurrence relation with same initial value is satisfied by the pseudo-Catalan number. Hence, h(n) = (2 * n – 2)! / (n – 1)!
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