How do you solve Sum of Subarray Ranges on LeetCode?

Sum of Subarray Ranges (LeetCode #2104) 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: Each element contributes to multiple subarrays as both maximum and minimum.

What is the brute force solution for Sum of Subarray Ranges?

The brute force approach for Sum of Subarray Ranges is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible subarrays and calculating their ranges by finding the maximum and minimum values for each subarray. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sum of Subarray Ranges?

Sum of Subarray Ranges uses the following concepts: Array, Stack, Monotonic Stack. The recommended patterns to study are: Monotonic Stack, Array.

What are the interview tips for Sum of Subarray Ranges?

Always think about how to reduce the number of operations — look for patterns. Practice using stacks for problems involving ranges or boundaries. Be prepared to explain your thought process clearly and justify your choice of algorithm.

What are common mistakes when solving Sum of Subarray Ranges?

Not considering all possible subarrays when calculating ranges. Failing to optimize the search for max/min values.

#2104

Sum of Subarray Ranges

Medium

ArrayStackMonotonic StackMonotonic StackArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a monotonic stack to efficiently calculate the contribution of each element as both a maximum and a minimum in all possible subarrays. This approach reduces the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Initialize a total sum variable to 0 and create two arrays to store the next and previous greater elements for each index.
2Step 2: Use a stack to find the next and previous greater elements for each element in the array.
3Step 3: For each element, calculate its contribution as a maximum and minimum based on its position in the subarrays.
4Step 4: Sum the contributions of all elements to get the final result.

solution.py22 lines

1def subArrayRanges(nums):
2    n = len(nums)
3    next_greater = [n] * n
4    prev_greater = [-1] * n
5    stack = []
6
7    for i in range(n):
8        while stack and nums[stack[-1]] < nums[i]:
9            next_greater[stack.pop()] = i
10        stack.append(i)
11
12    stack = []
13    for i in range(n - 1, -1, -1):
14        while stack and nums[stack[-1]] <= nums[i]:
15            prev_greater[stack.pop()] = i
16        stack.append(i)
17
18    total_sum = 0
19    for i in range(n):
20        total_sum += nums[i] * (next_greater[i] - i) * (i - prev_greater[i])
21
22    return total_sum

ℹ

Complexity note: This complexity is achieved because we traverse the array a constant number of times (once for finding next greater and once for previous greater), leading to linear time complexity.

1Each element contributes to multiple subarrays as both maximum and minimum.
2Using a monotonic stack helps efficiently find the next and previous greater elements.

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