Given a collection of n nuts, and a collection of n bolts, each arranged in an increasing order of size, give an O(n) time algorithm to check if there is a nut and a bolt that have the same size. You can assume that the sizes of the nuts and bolts are stored in the arrays NUTS[1::n] and BOLTS[1::n], respectively, where NUTS[1] < < NUTS[n] and BOLTS[1] < < BOLTS[n]. Note that you only need to report whether or not a match exists; you do not need to report all matches.

See explaination for the program code

Explanation:

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, output/return

}

else if nuts[i] < bolts[j] {

this means that size of nut is smaller than that of bolt and we should go to the next bigger nut, i.e., i+=1

}

else {

this means that size of bolt is smaller than that of nut and we should go to the next bigger bolt, i.e., j+=1

}

}

Since we go to each index in both the array only once, the algorithm take O(n) time.

1. the seasons of the world

well you could find somebody who doesn't want theirs and ask for it or possibly trick a dealer into giving it to you for free.