#3427

Easy

LeetCode Problem Workspace

Sum of Variable Length Subarrays

Calculate the total sum of all elements in subarrays defined for each index in an array, using the array plus prefix sum technique.

array prefix sum

Problem Statement

You are given an integer array nums of size n. For each index i where 0 <= i < n, define a subarray nums[start ... i] where start = max(0, i - nums[i]).

Return the total sum of all elements from the subarray defined for each index in the array. In simpler terms, you must compute the sum of elements from each subarray formed using the index-specific start and end indices.

Examples

Example 1

Input: nums = [2,3,1]

Output: 11

The total sum is 11. Hence, 11 is the output.

Example 2

Input: nums = [3,1,1,2]

Output: 13

The total sum is 13. Hence, 13 is the output.

Constraints

1 <= n == nums.length <= 100
1 <= nums[i] <= 1000

Solution Approach

Brute Force Approach

The simplest approach is to iterate through each index i and calculate the sum of elements in the subarray nums[start ... i], where start is calculated as max(0, i - nums[i]). While inefficient for large arrays, this solution works well given the constraints.

Prefix Sum Optimization

Using a prefix sum array, we can calculate the sum of elements from any subarray efficiently. By building a prefix sum array, we reduce the redundant calculations for each subarray sum, improving the overall time complexity.

Cumulative Sum Approach

Another approach involves maintaining a running total as you iterate through the array. By calculating the cumulative sum up to each index i, you can sum the required subarrays in linear time, ensuring that the solution remains efficient.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity can vary depending on the approach. The brute force approach runs in O(n^2) time, while using prefix sums can optimize this to O(n). The space complexity is O(n) due to the storage of the prefix sum array in the optimized solution.

What Interviewers Usually Probe

Candidate demonstrates understanding of prefix sums.
Candidate chooses an appropriate approach based on problem constraints.
Candidate avoids unnecessary recalculations in the solution.

Common Pitfalls or Variants

Common pitfalls

Forgetting to calculate the subarray's starting index properly.
Not considering the optimization potential with prefix sums.
Using inefficient algorithms for larger arrays without considering the constraints.

Follow-up variants

Given larger arrays, how can we optimize the brute force approach further?
What would happen if the array elements were negative? How would this affect the subarray calculations?
Can the solution be adjusted to return a different value, such as the maximum or minimum sum of the subarrays?

FAQ

What is the optimal solution for the Sum of Variable Length Subarrays problem?

The optimal solution uses prefix sums to calculate the sum of elements for each subarray efficiently, reducing redundant calculations.

How can brute force be applied to the Sum of Variable Length Subarrays?

Brute force involves calculating the sum of elements in each subarray independently, which can be slow but works due to small constraints.

What role does prefix sum play in solving this problem?

Prefix sum helps by allowing quick calculation of subarray sums, improving the solution from brute force's O(n^2) to O(n).

Can the Sum of Variable Length Subarrays be solved using dynamic programming?

While dynamic programming is not necessary, it can be used to store intermediate results to avoid redundant calculations, similar to prefix sums.

What are the time and space complexities of the optimal solution?

The optimal solution with prefix sums has a time complexity of O(n) and a space complexity of O(n), where n is the length of the array.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def subarraySum(self, nums: List[int]) -> int:
        s = list(accumulate(nums, initial=0))
        return sum(s[i + 1] - s[max(0, i - x)] for i, x in enumerate(nums))

Continue Practicing

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