Asked in: SALESFORCE
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All Testcases Passed ✔
Solution
def logTransferReceiver(logReceiver, k):
# Write your code here
log_starting_point = 0
last = 0
// ... rest of solution available after purchaseExplanation
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To solve this problem, you need to understand how a sequence of deterministic transfers over time can be analyzed efficiently, especially when the number of time steps (k) is very large—up to 10^12. At first glance, you might think about simulating the transfer of the debug log for k seconds by simply following the mapping from one developer to another step-by-step. However, that would be far too slow due to the constraint that k can be as large as a trillion.
Let’s start by understanding the structure of the data. Each element in the array logReceiver represents a directed edge from one developer to another. So, if logReceiver[i] = j, it means that developer i + 1 will pass the log to developer j after one second. This forms a graph of sorts, but it’s a special kind of graph—a functional graph—where each node has exactly one outgoing edge. That property is important because it means the graph is composed of one or more cycles and trees leading into those cycles.
Since developer 1 always starts with the log, you can follow the transfer path from developer 1. Because each developer always sends the log to exactly one other developer, and each transfer happens deterministically every second, the path the log follows is fixed. However, because each developer can appear only once in a transfer chain until a cycle is reached, eventually you will start to revisit developers. That is, the sequence will begin to repeat. When that happens, you’ve entered a cycle.
Here’s where the efficient thinking begins. Because the sequence will fall into a loop (a cycle), you don’t need to simulate all k steps. Instead, you only need to know two things: the length of the path before the cycle starts (this is often called the pre-cycle path), and the length of the cycle itself. Once you know those, you can reduce the problem by modding k appropriately.
So your thinking should progress in the following direction:
1. Simulate the transfers from developer 1, but not for k steps—only until you either reach a previously visited developer (i.e., a cycle), or until you hit k steps (whichever is smaller). Keep track of the order in which developers are visited.
2. As you simulate, maintain a mapping or record of the step (second) at which each developer is first visited. The moment you encounter a developer that has already been visited, you’ve identified the start of the cycle. From there, you can calculate the cycle length.
3. With the pre-cycle length and the cycle length known, determine whether k falls within the pre-cycle. If k is less than the length of the pre-cycle, you can directly return the developer who received the log at second k.
4. If k is larger than or equal to the pre-cycle length, subtract the pre-cycle length from k, and then use modulo arithmetic with the cycle length to find the final position in the cycle.
This approach brings your complexity down significantly. Instead of attempting to simulate all k transfers (which would be infeasible), you only simulate up to n steps to detect the cycle. That’s acceptable because even in the worst case, with n = 2 * 10^5, simulating n steps is feasible.
It’s also helpful to visualize a small example to cement your understanding. Let’s say you have logReceiver = [2, 3, 4, 2]. That means:
- Developer 1 passes to 2
- 2 passes to 3
- 3 passes to 4
- 4 passes to 2
Here, starting from 1, you get: 1 → 2 → 3 → 4 → 2 → 3 → 4 → ...
So the cycle is 2 → 3 → 4, and the pre-cycle is just the single transfer from 1 to 2. This means:
- At t = 0: Developer 1 has the log
- t = 1: Developer 2
- t = 2: Developer 3
- t = 3: Developer 4
- t = 4: Developer 2
- t = 5: Developer 3
- etc.
With this understanding, if you are asked where the log is after, say, k = 10 seconds, you note that 1 second was the pre-cycle and the cycle length is 3. Subtract 1 from 10 to get 9, then compute 9 mod 3 = 0. That means you are at the beginning of the cycle, which corresponds to Developer 2.
So your approach should not involve brute-force simulation for k seconds, but rather smart identification of the repeating cycle and modular arithmetic to jump directly to the answer. Keep track of the visit order, map developer IDs to the time they were visited, and once a repetition is found, use it to compute the final answer in constant time.
By analyzing the cycle behavior and utilizing simple mathematical tools like modulo, you can answer even very large queries efficiently.
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