Finding Matching Nuts and Bolts in O(n) Time
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, how can we find a nut and a bolt that have the same size? What is the time complexity of the algorithm?
Solution: Matching Nuts and Bolts
To check if there is a nut and a bolt that have the same size, we can use 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, output/return
} else if nuts[i] < bolts[j] {
This means the size of nut is lesser than that of bolt, so we move to the next bigger nut, i.e., i+=1
} else {
This means the size of bolt is lesser than that of nut, so we move to the next bigger bolt, i.e., j+=1
}}
Since we iterate through each index in both arrays only once, the algorithm runs in O(n) time complexity.
Explanation
When given a collection of nuts and bolts sorted in increasing order of sizes, we can efficiently find a nut and a bolt that have the same size using a two-pointer approach. By comparing the sizes of nuts and bolts iteratively, we can identify the matching pair with a time complexity of O(n).
The algorithm involves keeping two iterators, one for the nuts array and another for the bolts array. By comparing the sizes at the current positions of the iterators, we can determine the relationship between the sizes and adjust the iterators accordingly. If a match is found, we can immediately output or return the pair without the need to continue searching for additional matches.
By iterating through each index in both the nuts and bolts arrays only once, the algorithm ensures that the time complexity is linear, making it efficient for checking matching sizes in a large collection of nuts and bolts. This method optimizes the process by utilizing the sorted order of the arrays to quickly identify matching pairs while maintaining a linear runtime.