Algorithms : Divide and Conquer

A divide and conquer algorithm repeatedly reduces an instance of a problem to one or more smaller instances of the same problem until the instances are small enough to solve easily.

Equal number of nuts and bolts of various sizes are given. There is a match for every bolt in the set of the nuts. Comparisom of a bolt with a bolt and a nut with a nut is prohibited. How to make ut bolt pairs in shortest possible time?

We can use divide and conquer here. We cannot sort only nuts ( only bolts ) according to size because nut-nut (bolt-bolt ) comparison is prohibited.

So we try to match a nut with bolts. Three conditions arise :
a. Its a match.
b. Bolt is smaller.
c. Bolt is bigger.

Now we can divide bolts in two parts: smaller ones and bigger ones. The matching nut-bolt (pivot elements) are the first match.

Now try to match the pivot bolt with remaining nuts. Two conditions arise :
a. Nut is smaller.
b. Nut is bigger.

Now we can divide nuts in two parts: smaller ones and the bigger ones.

The set of smaller nuts have a match in the set of smaller bolts and vice versa.

So after ~2n (assuming n nuts and n bolts) comparisons the bigger problem is divided into 2 smaller problems of half size (~ n/2).

Hence the time complexity is:

T(n)= T(n/2)+ O(n) ~ O(n lg n).