Asked in: SHARECHAT
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All Testcases Passed ✔
Solution
from functools import lru_cache
def oneBlock(N, arr):
@lru_cache(None)
def solve(i, flag):
// ... rest of solution available after purchaseExplanation
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To solve this problem, the main goal is to find the number of ways to split the given array into continuous segments (blocks), where each segment contains exactly one resource with value 1. Such a segment is called a “One block.” If it’s impossible to split the array into such One blocks, we return 0.
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### Understanding the Problem
You have an array `Arr` of length `N`. Each element is a resource value, either 0 or 1 (though not explicitly stated, the problem implies only these two since One block requires exactly one 1). You want to split the array into contiguous subarrays so that each subarray contains exactly one occurrence of the value 1.
For example, if the array is `[0, 1, 0, 1, 0]`, possible One blocks are subarrays like `[0, 1, 0]` or `[1]`. The whole array can be split into blocks `[0, 1, 0]` and `[1, 0]`, each containing exactly one ‘1’.
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### Key Points and Constraints
- Every block must have exactly one '1'.
- Blocks are continuous subarrays.
- You cannot reorder elements; splitting only.
- The number of resources `N` can be up to 100, which suggests that an O(N²) solution is feasible.
- If no valid splitting exists, return 0.
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### Approach Overview
1. **Identify positions of all the 1s in the array.**
Since every block must contain exactly one ‘1’, these 1s are essentially anchors around which blocks are formed.
For instance, if your array is `[0, 1, 0, 1, 0, 1]`, the indices of 1s are `[1, 3, 5]` (0-based indexing). Each block will include exactly one of these indices.
2. **Divide the problem into segments between these 1s**
Each block must cover exactly one ‘1’, so the blocks are formed between the positions of these 1s.
The first block must start from the beginning of the array and end at or after the first ‘1’ (but before the second ‘1’). The last block must start at or after the last ‘1’ and continue to the end.
3. **Count the possible ways to form blocks between two consecutive 1s**
For two consecutive 1s at positions `i` and `j`, the subarray that forms a block with the 1 at position `i` can extend from the previous 1's position + 1 up to just before position `j`.
Since the array is contiguous, the number of ways to form a block between these two ‘1’s depends on the number of zeros between them.
The more zeros between two 1s, the more ways you can extend the block containing the first ‘1’.
4. **Calculate the total number of ways by multiplying possibilities**
For each pair of consecutive 1s, compute the number of possible block divisions, which is related to the count of zeros between them plus one.
Multiply the number of ways for each such pair to get the total number of ways to split the entire array into One blocks.
5. **Edge cases**
- If there are no 1s in the array, return 0 since you cannot form any One block.
- If there is only one ‘1’, there is only one way to form the block (the entire array is one block).
- If the array starts or ends with zeros, that doesn’t affect the count except in how many zeros are between the 1s.
- Consecutive 1s (e.g., `[1, 1]`) mean the block with the first ‘1’ ends immediately before the second ‘1’, so only one way for that pair.
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### Detailed Thought Process
- Extract the indices of all 1s and store them.
- Iterate through the array to count zeros between consecutive 1s.
- For each gap between the ith and (i+1)th ‘1’, calculate the number of zeros between them.
- Each zero in the gap allows an additional way to break the block.
- Total ways = product over all gaps of (number_of_zeros_in_gap + 1).
- This multiplication arises because for each gap, you can choose the block boundary in multiple ways independently.
- The final answer is this product.
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### Why does this work?
Because blocks must be continuous and contain exactly one ‘1’, the only variability is in how many zeros you include with the current ‘1’. Zeros are “free” to be included either in the current block or the next one (except that the next block must start at the next ‘1’). So the possible splits come from how you assign these zeros between blocks.
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### Complexity
- Finding indices of 1s is O(N).
- Calculating zeros between each pair of 1s is O(N).
- Final multiplication is O(number_of_1s), which is at most N.
- Total complexity is O(N), efficient for N up to 100 or even larger.
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### Summary
The problem reduces to counting the ways to distribute zeros between consecutive 1s while ensuring exactly one ‘1’ per block. The approach uses simple array traversal, position tracking of ‘1’s, and combinatorial multiplication to get the final count.
If no 1 is present, or the input is invalid, return 0.
This approach combines array analysis with combinatorial logic, avoiding brute force enumeration of all partitions and instead using the structure of the problem for efficient counting.
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