Suppose we have two arrays A[n] and b[n], the goal is to compare every element in A to elements in B. Then return a list result[n] that records the number of each element in A that is larger than the elements in B.
For example,
A = [38, 24, 43, 3], B = [9, 82, 10, 11]
Since 38 is larger than 9, 10 and 11, so result[0] is 3. Then result is [3, 3, 3, 0].
It will be the best if you can provide some pseudocode.
Thank you.
You can perform the above algorithm in O(nlogn) complexity where n is the length of array A and array B as given in the question.
1. Sort both the arrays A and B, this will take O(nlogn) time complexity.
2. Take two pointers i and j, initialize both of them to 0. we will use i for array A and j for B.
3. Create a result array res of size n.
4. Start a while loop
while(i<n && j<n) {
if(A[i] > B[j]) {
j++;
} else {
res[i] = j+1;
i++;
}
}
5. while(i<n) {
res[i] = n;
}
This step is for the case where all elements in A are bigger than all elements in B.
At the end you will have res array ready with the answer.
Overall time complexity - O(nlogn).
Hope this helps!
Both lists need to be sorted ascending for this to work.
Sorting costs O(log n). And big-O arithmetic means doing it twice is still O(n log n). I'm going to assume they are already sorted. The remaining work below doesn't affect the big-O cost.
Have an index to the B array called indexB, value zero (my pseudocode will use zero-based indexing). And indexA for A also starting at zero.
indexA=0
For each indexB from 0 to B.Length-1
While indexA < A.Length and the value at `A[indexA]` is less than or equal to the value at `B[indexB]`
Set the `result[indexA]` to be `indexB`
Increment `indexA`
Endwhile
Endfor
After that, all of the remaining items in the result from indexA onward are bigger than all the items in B, so set the rest of them to B.Length.
