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Problem solution · Python

Find Maximum Non-decreasing Array Length

Find Maximum Non-decreasing Array Length: a Python solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
26 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

Dynamic programming

For Find Maximum Non-decreasing Array Length, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 26 lines of Python from the credited upstream file 2945-2.py.
  • The implementation visibly relies on sequence storage, cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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.

Open problem ↗Original source ↗
Full codeFind Maximum Non-decreasing Array Length · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def findMaximumLength(self, nums: list[int]) -> int:    n = len(nums)    # prefix[i] := the sum of the first i nums    prefix = list(itertools.accumulate(nums, initial=0))    # dp[i] := the maximum number of elements in the increasing    # sequence after processing the first i nums    dp = [0] * (n + 1)    # bestLeft[i] := the index l s.t. merging nums[l..i) is the    # optimal strategy among processing the first i nums    bestLeft = [0] * (n + 2)     for i in range(1, n + 1):      bestLeft[i] = max(bestLeft[i], bestLeft[i - 1])      # When merging nums[l, i), consider the next segment as [i, r).      # Find the minimum `r` where sum(nums[l, i)) <= sum(nums[i, r)).      # Equivalently, prefix[i] - prefix[l] <= prefix[r] - prefix[i].      #            => prefix[r] >= prefix[i] * 2 - prefix[l]      # Therefore, we can binary search `prefix` to find the minimum `r`.      l = bestLeft[i]      r = bisect.bisect_left(prefix, 2 * prefix[i] - prefix[l])      dp[i] = dp[l] + 1      bestLeft[r] = i     return dp[n]

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