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Solution
def subarraySum(arr):
ans = 0
left = 0
n = len(arr)
// ... rest of solution available after purchaseExplanation
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To approach the problem of calculating the sum of all subarrays of a given array of integers, it's essential to first understand what a subarray is and how it contributes to the total sum.
A subarray is a contiguous segment of the array. For an array with `n` elements, there are many subarrays of varying lengths: from 1-element subarrays up to the entire array itself. For instance, if your array is `[a, b, c]`, the subarrays are:
- `[a]`
- `[b]`
- `[c]`
- `[a, b]`
- `[b, c]`
- `[a, b, c]`
Each of these subarrays contributes to the final sum by the sum of its elements. So, your goal is to compute the total of all these contributions. A naive method would involve generating all possible subarrays, summing their elements, and adding that to a running total. But this approach becomes impractical with large arrays, as it would require iterating over potentially millions or billions of subarrays, leading to time inefficiency.
To think in the right direction, consider instead how often each element appears in various subarrays. This leads to a more mathematical approach that avoids iterating over all subarrays directly.
Here’s the key insight: for a given element at index `i` in the array, it appears in many subarrays. In fact, it appears in all subarrays that start at or before index `i` and end at or after index `i`. You can calculate the number of such subarrays using combinatorial reasoning.
Let’s say you have an array of length `n`. For an element at position `i` (using 0-based indexing), there are:
- `(i + 1)` choices for where a subarray can start that includes this element (it can start at any index from 0 to i)
- `(n - i)` choices for where the subarray can end (it can end at any index from i to n - 1)
This means that the element at index `i` appears in exactly `(i + 1) * (n - i)` subarrays. This is a crucial realization because instead of generating all subarrays, you can compute the contribution of each individual element by multiplying the value of the element by the number of subarrays it appears in.
So the total sum of all subarrays can be computed by summing, for each index `i`, the product: `arr[i] * (i + 1) * (n - i)`
This approach ensures that you process each element only once, and the entire computation can be done in a single loop through the array. This is critical because the constraints allow arrays with up to 200,000 elements, and an inefficient approach could easily lead to timeouts or excessive resource consumption.
Let’s consider a small example to solidify the logic:
Take the array `[4, 5, 6]`
- For element 4 at index 0: It appears in 1 * 3 = 3 subarrays → contributes 4 * 3 = 12
- For element 5 at index 1: It appears in 2 * 2 = 4 subarrays → contributes 5 * 4 = 20
- For element 6 at index 2: It appears in 3 * 1 = 3 subarrays → contributes 6 * 3 = 18
Total = 12 + 20 + 18 = 50, which matches the manual subarray sum.
This conceptual model significantly reduces computational complexity. Instead of building subarrays and computing their sums directly, which would be a nested operation (resulting in O(n^2) time), this approach works in linear time — O(n) — which is acceptable given the constraint of up to 200,000 elements.
Another way to think about this approach is to focus on the *impact* each element has on the final result. Rather than building out subarrays, count how many times each element impacts the total, and sum those contributions directly.
When building your function, all you need to do is:
- Loop over the array with a single pass.
- For each element, compute its contribution based on its index and array length.
- Add that contribution to a running total.
This keeps the memory usage minimal, avoids complex data structures or intermediate storage of subarrays, and scales efficiently with input size.
In summary, shift your thinking from "building subarrays" to "counting element participation." This reframing leads you to an efficient and elegant solution that leverages mathematical insight over brute force iteration. It also aligns perfectly with the problem constraints and expected performance.
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