How do you solve Maximum Subarray Min-Product on LeetCode?

Maximum Subarray Min-Product (LeetCode #1856) 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: Using prefix sums allows for quick calculation of subarray sums.

What is the brute force solution for Maximum Subarray Min-Product?

The brute force approach for Maximum Subarray Min-Product is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible subarray to calculate its min-product. This is straightforward but inefficient for larger arrays due to its quadratic time complexity. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Subarray Min-Product?

Maximum Subarray Min-Product uses the following concepts: Array, Stack, Monotonic Stack, Prefix Sum. The recommended patterns to study are: Monotonic Stack, Prefix Sum, Array.

What are the interview tips for Maximum Subarray Min-Product?

Always discuss your thought process and approach before coding. Consider edge cases, such as arrays with all identical elements or very small sizes. Practice using stacks and prefix sums in different contexts to become comfortable with these patterns.

What are common mistakes when solving Maximum Subarray Min-Product?

Not considering all possible subarrays when implementing the brute force solution. Failing to manage stack operations correctly in the optimal solution.

#1856

Maximum Subarray Min-Product

Medium

ArrayStackMonotonic StackPrefix SumMonotonic StackPrefix SumArray

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 maximum min-product for each element in the array. This approach reduces the time complexity significantly by avoiding the need to check all subarrays explicitly.

⚙️

Algorithm

4 steps

1Step 1: Create a prefix sum array to store the cumulative sums of nums.
2Step 2: Use a monotonic stack to find the next and previous smaller elements for each element in nums.
3Step 3: For each element, calculate the min-product using the prefix sums and the identified boundaries.
4Step 4: Keep track of the maximum min-product found and return it modulo 10^9 + 7.

solution.py30 lines

1# Full working Python code
2
3def maxSubarrayMinProduct(nums):
4    MOD = 10**9 + 7
5    n = len(nums)
6    prefix_sum = [0] * (n + 1)
7    for i in range(n):
8        prefix_sum[i + 1] = prefix_sum[i] + nums[i]
9
10    stack = []
11    max_product = 0
12    for i in range(n):
13        while stack and nums[stack[-1]] > nums[i]:
14            j = stack.pop()
15            left = stack[-1] if stack else -1
16            right = i
17            min_val = nums[j]
18            total_sum = prefix_sum[right] - prefix_sum[left + 1]
19            max_product = max(max_product, min_val * total_sum)
20        stack.append(i)
21
22    while stack:
23        j = stack.pop()
24        left = stack[-1] if stack else -1
25        right = n
26        min_val = nums[j]
27        total_sum = prefix_sum[right] - prefix_sum[left + 1]
28        max_product = max(max_product, min_val * total_sum)
29
30    return max_product % MOD

ℹ

Complexity note: The time complexity is O(n) because we traverse the array and use the stack to find boundaries efficiently. The space complexity is O(n) due to the prefix sum array and the stack.

1Using prefix sums allows for quick calculation of subarray sums.
2Monotonic stacks help efficiently find boundaries for minimum values in subarrays.

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