Problem solution · Python

Maximum and Minimum Sums of at Most Size K Subarrays

Maximum and Minimum Sums of at Most Size K Subarrays: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
50 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Maximum and Minimum Sums of at Most Size K Subarrays, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 50 lines of Python from the credited upstream file 3430.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Full codeMaximum and Minimum Sums of at Most Size K Subarrays · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  # Similar to 2104. Sum of Subarray Ranges  def minMaxSubarraySum(self, nums: list[int], k: int) -> int:    prevGt, nextGt = self._getPrevNext(nums, operator.lt)    prevLt, nextLt = self._getPrevNext(nums, operator.gt)    return (self._subarraySum(nums, prevGt, nextGt, k) +            self._subarraySum(nums, prevLt, nextLt, k))   def _subarraySum(      self,      nums: list[int],      prev: list[int],      next: list[int],      k: int  ) -> int:    """    Returns the sum of all subarrays with a size <= k, The `prev` and `next`    arrays are used to store the indices of the nearest numbers that are    smaller or larger than the current number, respectively.    """    res = 0    for i, num in enumerate(nums):      l = min(i - prev[i], k)      r = min(next[i] - i, k)      extra = max(0, l + r - 1 - k)      res += num * (l * r - extra * (extra + 1) // 2)    return res   def _getPrevNext(      self,      nums: list[int],      op: callable  ) -> tuple[list[int], list[int]]:    """    Returns `prev` and `next`, that store the indices of the nearest numbers    that are smaller or larger than the current number depending on `op`.    """    n = len(nums)    prev = [-1] * n    next = [n] * n    stack = []    for i, num in enumerate(nums):      while stack and op(nums[stack[-1]], num):        index = stack.pop()        next[index] = i      if stack:        prev[i] = stack[-1]      stack.append(i)    return prev, next 

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