Generating Login Codes

In a software company, each employee's login process involves two arrays: initialLogin of size n and standardLogin of size m.
The security software transforms these arrays by repeatedly performing an operation:
Select any subsegment of either array and replace it with the sum of its elements.
For example, the array [1, 5, 6, 8, 2] can be transformed into [12, 8, 2] by replacing the subsegment [1, 5, 6] with [12].
The goal is to maximize the length of equal arrays after performing the operations any number of times on both initialLogin and standardLogin. The login code is the maximum possible length of these equal arrays. If the arrays cannot be made equal through the operations, the initialLogin is considered invalid, and the result should be -1.
Determine the login code based on the provided initialLogin and standardLogin, or return -1 if initialLogin is invalid.

Function
getLoginCodes(initialLogin: int[], standardLogin: int[]) → int
Complete the function getLoginCodes in the editor with the following parameters:
initialLogin[n]: the initial array
standardLogin[m]: the standard array used by the security software
Returns int: the login code

Examples
Example 1
initialLogin = [2, 4, 3, 7, 10]
standardLogin = [6, 5, 5, 10]
return = 3
n = 5, initialLogin = [2, 4, 3, 7, 10], m = 4, and standardLogin = [6, 5, 5, 10].
An optimal sequence is:
Operation NumberOperationinitialLogin After OperationstandardLogin After Operation
1Replace the subsegment [3, 7] with [10] in initialLogin.[2, 4, 10, 10][6, 5, 5, 10]
2Replace the subsegment [5, 5] with [10] in standardLogin.[2, 4, 10, 10][6, 10, 10]
3Replace the subsegment [2, 4] with [6] in initialLogin.[6, 10, 10][6, 10, 10]
Return the maximum possible length, 3.

Merging subsegments means each final element is a contiguous block sum. Two final arrays are equal only if their block boundaries land on the same prefix sums. So the answer is the count of shared prefix sums between the two arrays, including the total. Use two pointers: walk both arrays, keep running sums a and b, advance whichever is smaller, and when they're equal, increment the count and move both on. If the total sums differ, return -1 right away. Check that first. In the example, prefix sums of [2,4,3,7,10] are 2,6,9,16,26 and of [6,5,5,10] are 6,11,16,26. Shared: 6, 16, 26, so 3. The common pitfall is trying to simulate merges or run DP, which is overkill. Another is forgetting that all elements are assumed positive, which is what makes greedy matching valid. It's O(n+m) time and O(1) space. If you freeze on the reduction during the live OA, StealthCoder is the hedge that surfaces this approach.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.