How do you solve Sum of Variable Length Subarrays on LeetCode?

Sum of Variable Length Subarrays (LeetCode #3427) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to calculate subarray sums efficiently is crucial.

What is the brute force solution for Sum of Variable Length Subarrays?

The brute force approach for Sum of Variable Length Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach calculates the sum of subarrays for each index by iterating through the possible starting points. This method is straightforward but inefficient for larger arrays due to its nested loops. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sum of Variable Length Subarrays?

Sum of Variable Length Subarrays uses the following concepts: Array, Prefix Sum. The recommended patterns to study are: Prefix Sum, Array.

What are the interview tips for Sum of Variable Length Subarrays?

Always think about how to optimize your solution after the brute force approach. Be clear about your thought process when explaining your solution. Practice dry runs of your code to ensure correctness.

What are common mistakes when solving Sum of Variable Length Subarrays?

Not considering the boundaries when calculating the starting index. Forgetting to initialize the prefix sum array correctly.

#3427

Sum of Variable Length Subarrays

Easy

ArrayPrefix SumPrefix SumArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach uses a prefix sum array to efficiently calculate the sum of subarrays without repeatedly summing elements. This reduces the time complexity significantly.

⚙️

Algorithm

7 steps

1Step 1: Create a prefix sum array where prefix[i] is the sum of elements from nums[0] to nums[i].
2Step 2: Initialize total_sum to 0.
3Step 3: Loop through each index i from 0 to n-1.
4Step 4: Determine the starting point start = max(0, i - nums[i]).
5Step 5: Calculate the subarray sum using the prefix sum array: sum = prefix[i] - (start > 0 ? prefix[start - 1] : 0).
6Step 6: Add the calculated sum to total_sum.
7Step 7: Return total_sum after processing all indices.

solution.py11 lines

1def sum_of_variable_length_subarrays(nums):
2    n = len(nums)
3    prefix = [0] * n
4    prefix[0] = nums[0]
5    for i in range(1, n):
6        prefix[i] = prefix[i - 1] + nums[i]
7    total_sum = 0
8    for i in range(n):
9        start = max(0, i - nums[i])
10        total_sum += prefix[i] - (prefix[start - 1] if start > 0 else 0)
11    return total_sum

ℹ

Complexity note: The time complexity is O(n) because we only loop through the array a constant number of times. The space complexity is O(n) due to the prefix sum array.

1Understanding how to calculate subarray sums efficiently is crucial.
2Prefix sums can significantly reduce the time complexity of problems involving cumulative sums.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.