What is the brute force solution for Maximum and Minimum Sums of at Most Size K Subarrays?

The brute force approach for Maximum and Minimum Sums of at Most Size K Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subarrays of size up to k and calculating their minimum and maximum values. 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 Maximum and Minimum Sums of at Most Size K Subarrays?

Maximum and Minimum Sums of at Most Size K Subarrays uses the following concepts: Array, Math, Stack, Monotonic Stack. The recommended patterns to study are: Monotonic Stack, Sliding Window.

What are the interview tips for Maximum and Minimum Sums of at Most Size K Subarrays?

Always clarify the constraints and edge cases before jumping into coding. Explain your thought process while coding to demonstrate your understanding. Practice problems involving stacks and subarrays to build familiarity with patterns.

What are common mistakes when solving Maximum and Minimum Sums of at Most Size K Subarrays?

Not handling edge cases where k is equal to the length of the array. Failing to correctly implement the logic for calculating contributions using stacks.

#3430

Maximum and Minimum Sums of at Most Size K Subarrays

Hard

ArrayMathStackMonotonic StackMonotonic StackSliding Window

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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 the maximum and minimum in all valid subarrays. This reduces the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Use two monotonic stacks to find the next and previous greater and smaller elements for each element in the array.
2Step 2: For each element, calculate how many subarrays it contributes as the maximum and minimum based on the indices found in the stacks.
3Step 3: Multiply the contribution by the element's value and accumulate the results.

solution.py44 lines

1# Full working Python code
2
3def max_min_sum(nums, k):
4    n = len(nums)
5    total_sum = 0
6    max_contrib = 0
7    min_contrib = 0
8
9    # Monotonic stacks for max
10    max_stack = []
11    for i in range(n):
12        while max_stack and nums[max_stack[-1]] < nums[i]:
13            j = max_stack.pop()
14            left = j - (max_stack[-1] if max_stack else -1)
15            right = i - j
16            max_contrib += nums[j] * left * right
17        max_stack.append(i)
18
19    while max_stack:
20        j = max_stack.pop()
21        left = j - (max_stack[-1] if max_stack else -1)
22        right = n - j
23        max_contrib += nums[j] * left * right
24
25    # Monotonic stacks for min
26    min_stack = []
27    for i in range(n):
28        while min_stack and nums[min_stack[-1]] > nums[i]:
29            j = min_stack.pop()
30            left = j - (min_stack[-1] if min_stack else -1)
31            right = i - j
32            min_contrib += nums[j] * left * right
33        min_stack.append(i)
34
35    while min_stack:
36        j = min_stack.pop()
37        left = j - (min_stack[-1] if min_stack else -1)
38        right = n - j
39        min_contrib += nums[j] * left * right
40
41    return max_contrib - min_contrib
42
43# Example usage
44print(max_min_sum([1, 2, 3], 2))  # Output: 20

ℹ

Complexity note: The time complexity is O(n) because each element is pushed and popped from the stack at most once. The space complexity is O(n) due to the stacks used to store indices.

1Using monotonic stacks helps efficiently calculate contributions of elements as maximums and minimums.
2Understanding the contribution of each element in terms of its position and neighboring elements is crucial.

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