Ethan Klein
I am a professional Digital Marketer in a technology company. I have More than 10 years of experience as a Digital Marketing Manager. In our company, we provide that kind of services Business Analysis, Product Management, Solution Design & Development, Team Management, Sales/Marketing Teams, CRM systems (Salesforce, SugarCRM, Podio, Zoho), marketing consultancy services, digital marketing and branding services, BPO and Virtual Staffing.
To solve this problem, we can follow the following steps:
- Ethan Klein March 31, 2023Create a hashmap of the values in array A, where the keys are the values and the values are the indices.
Sort array B in ascending order.
Iterate through each value b in B.
If b is greater than the maximum value in A, we can skip it since it won't have any effect.
If b is less than or equal to the minimum value in A, we can also skip it since it won't have any effect.
Find the index i of the smallest value in A that is greater than or equal to b using binary search.
Increment the value of A[i-1] by 1.
If A[i-1] equals A[i], remove A[i-1] from the hashmap.
Append the updated A to the output array Z.
Return Z.
The worst-case time complexity of this algorithm is O(M + N + H), where H is the maximum value in array B. This is because we need to iterate through all the values in A and B and perform binary search for each value in B. The worst-case space complexity is O(M), as we only need to store the hashmap of values in A.
Here is the Python code for this algorithm:
def generate_output_array(A, B):
value_index_map = {value: index for index, value in enumerate(A)}
B.sort()
output = []
for b in B:
if b > A[-1] or b <= A[0]:
continue
i = bisect_left(A, b)
A[i-1] += 1
if A[i-1] == A[i]:
del value_index_map[A[i-1]]
output = A[:]
return output