Asked in: GOLDMANSACHS
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All Testcases Passed ✔
Solution
#!/bin/python3
import math
import os
// ... rest of solution available after purchaseExplanation
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To solve this problem, you should start by visualizing the toy distribution process as a circular sequence. Each kid has a unique ID from 1 to N, and the kids are seated in a circle. This circular arrangement is key to understanding how the toys are handed out: after the last kid (with ID N), the next toy goes to the first kid (with ID 1), and the cycle continues until all T toys are given.
Now, consider that the host picks a starting kid with ID D. The first toy goes to kid D, the second to D+1, and so on. But since the arrangement is circular, any time the count exceeds N (the highest ID), it wraps back around to 1. This is a classic case of circular iteration, and it can be understood in terms of modular arithmetic.
Think about what happens when you distribute T toys starting from position D. Each toy is handed to the next kid in order, wrapping around at the end of the list. The goal is to determine who gets the last toy, i.e., the T-th one. The simplest approach is to simulate this progression, counting forward one by one, while wrapping around when the count goes past N.
However, instead of simulating the entire process toy by toy, you can think about it mathematically. What you need is to compute the position in the circle after taking T - 1 steps from the starting point D. The reason it's T - 1 is because the first toy goes to D itself, and we’re interested in where the last toy ends up — which would be the result of moving forward T - 1 steps.
Now, this step-counting process loops back to the beginning once it goes beyond N, so you can use modular arithmetic to determine where you’ll land. But be careful: since kid IDs start from 1 (not 0), you need to adjust your thinking slightly to account for 1-based indexing. With zero-based indexing, modulo operations are straightforward. But when you start counting from 1, you must shift the numbers accordingly to maintain correctness.
More formally, consider the circle to be a number line of length N, starting at position D. If you add T - 1 steps to D and then wrap around using modulus N, you’ll arrive at the final recipient’s ID. One key detail here is ensuring that if your modulo operation results in 0 (which can happen in 1-based systems), it should actually represent the last kid (with ID N), not zero. That’s a small but important caveat.
In the sample input example, you have N = 5 kids, T = 2 toys, and you start from D = 1. So:
- First toy goes to kid 1.
- Second (and last) toy goes to kid 2.
Thus, the output is 2. If you look closely, you are really adding T - 1 = 1 to the starting position, then checking where you land within a circular list of size N.
So, the generalized approach is:
1. Treat the distribution process as moving (T - 1) steps from the starting kid D.
2. Apply circular logic using modulo arithmetic to wrap around the kid IDs once you exceed N.
3. Adjust for 1-based indexing, especially in how the modulo operation is interpreted. If your final result after modulo is 0, then it means the last kid (ID N) gets the toy.
This reasoning removes the need for simulating the entire process step-by-step, which is inefficient for large T or N. Instead, by reducing the problem to a circular counting problem, you get a much faster and cleaner solution.
To summarize:
- Think of kids seated in a circle numbered 1 to N.
- Start distributing from kid D.
- Each toy goes to the next kid in sequence, wrapping back to 1 after reaching N.
- Move forward T - 1 positions from D to find who receives the last toy.
- Use modular arithmetic to handle the circular nature and adjust for 1-based indexing.
By following this reasoning, you can determine the final recipient of the damaged toy with both accuracy and efficiency, regardless of the values of N, T, and D.
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