Asked in: UKG
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All Testcases Passed ✔
Solution
from functools import lru_cache
from sys import setrecursionlimit
setrecursionlimit(10**6)
// ... rest of solution available after purchaseExplanation
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To solve this problem, we need to deeply understand the nature of the operation and how it affects the elements in the array. The core operation allows us to choose a positive integer x and subtract 1 from every element in the array that is greater than or equal to x. This operation can be performed any number of times, with possibly different values of x each time. Our task is to determine how many unique final arrays we can obtain after applying this operation any number of times, including zero.
Let’s break this down by first understanding what the operation actually does to the array. Choosing a value of x and performing the operation reduces certain elements — those that are at least x — by 1. Over time, as we repeat such operations with different values of x, various elements in the array can decrease, but never increase. Eventually, all elements in the array can be brought down to zero, but the path to zero — meaning the order and choice of x values — determines the shape of the resulting array before reaching all zeros.
Now, consider that at any point, you can only decrease elements that are above or equal to the x you choose. So, when you apply this operation with x = 1, all elements greater than or equal to 1 are reduced. If you then apply x = 2, all values that are now at least 2 get reduced again. The key is to recognize that the same value can be affected by multiple operations if it remains above or equal to the chosen x at each step.
To reason about the number of different final arrays possible, we can think of it in terms of a tree or graph where each node represents a state of the array and edges represent a valid operation that transitions from one array to another. However, traversing this graph directly is computationally impossible due to the exponential size of the state space. Instead, we need to observe patterns and structure that allow us to count without enumeration.
A powerful insight comes from considering the maximum values of the array elements and how they constrain the number of ways the values can be decreased. Take each element independently and imagine its own "path" to zero. For example, if an element is 3, then through a sequence of operations it can be reduced to 0, and at each step it can be reduced depending on what x values are chosen. The combinations of these reductions across different elements define the different final arrays.
So, we need to consider how many distinct combinations of values can exist where each element is less than or equal to its original value, and where the values are reduced in such a way that the transitions are consistent with the operation’s rule. But we must avoid overcounting duplicate final arrays that may result from different sequences of operations leading to the same array.
This leads to an important conceptual simplification: instead of thinking about the order of operations, focus on the final values each element can take. Since elements can only be reduced and cannot be increased, each element’s value in a final array must be less than or equal to its original value.
Now, consider that if we sort the original array in non-decreasing order, and process it element by element, we can build up a count of how many different ways reductions can be applied. More specifically, think about unique values in the array — every time a new value appears that hasn't been seen before, it opens up a new dimension of choice in how the operation can be applied. For instance, if the array contains values [2, 3, 0, 2], the unique values are {0, 2, 3}. Each unique value allows a kind of "branching" in how it can be reduced compared to others.
So the number of different final arrays is directly tied to the number of unique values in the array. Each unique value represents a decision point: do we reduce the elements at that level or not? Since we can make a binary decision (reduce or don’t reduce) for each such unique value (except zero, which can't be reduced further), the number of distinct final arrays is 2^(number of unique positive values). That’s because every combination of applying or skipping reduction at a specific level leads to a unique configuration of the array.
Finally, because the number of such combinations can be very large, we take the result modulo 10^9 + 7 as specified.
To summarize:
1. Understand the effect of the operation: it subtracts 1 from values ≥ x.
2. Recognize that the same final array can be reached through multiple operation sequences, but we only count unique final configurations.
3. Focus on the final possible values each element can take, not the sequence of operations.
4. Identify the unique positive values in the array, which represent levels at which decisions can be made.
5. Each unique positive value offers a binary choice: apply the operation or not.
6. Therefore, the total number of different final arrays is 2^(number of unique positive values), modulo 10^9 + 7.
This way, you can compute the number of different final arrays without simulating every operation or final state, which makes the solution efficient and scalable.
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