Efficient Algorithm for Matching Nuts and Bolts

How can we efficiently check if there is a nut and a bolt that have the same size?

Given a collection of n nuts and a collection of n bolts, arranged in an increasing order of size, is there a way to achieve this matching effectively?

Efficient Matching Algorithm

One efficient way to check if there is a nut and a bolt that have the same size is by using the following algorithm:

• Keep two iterators, i (for nuts array) and j (for bolts array).
• While i < n and j < n:
  • If nuts[i] == bolts[j], we have a case where sizes match, and we can output or return the result.
  • Else if nuts[i] < bolts[j], move to the next bigger nut by incrementing i (i.e., i+=1).
  • Else if bolts[i] > nuts[j], move to the next bigger bolt by incrementing j (i.e., j+=1).

Explanation

This algorithm works by comparing the sizes of nuts and bolts using two iterators. By iterating through the arrays of nuts and bolts simultaneously, we can efficiently find a matching pair without the need to report all matches. The algorithm stops as soon as it finds a single match.

Since the algorithm traverses each index of both arrays only once, it operates in O(n) time complexity. This means that the time taken to complete the matching process is proportional to the number of elements in the arrays, providing an efficient solution for matching nuts and bolts of the same size.