Def_pow_(self,b): if not then I’m sorry
Answer:
See explaination
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.
Answer:
phone !!
Explanation:
since it's a mobile app, that would apply to a handheld device !!
i hope this helps !!
the undo option is the right answer
