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Problem solution · Python

Sum of All Odd Length Subarrays

Sum of All Odd Length 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
22 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 Sum of All Odd Length 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

  • 22 lines of Python from the credited upstream file 1588.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.

Open problem ↗Original source ↗
Full codeSum of All Odd Length Subarrays · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def sumOddLengthSubarrays(self, arr: list[int]) -> int:    ans = 0    # Maintain two sums of subarrays ending in the previous index.    # Each time we meet a new number, we'll consider 'how many times' it should    # contribute to the newly built subarrays by calculating the number of    # previous even/odd-length subarrays.    prevEvenSum = 0  # the sum of even-length subarrays    prevOddSum = 0  # the sum of odd-length subarrays     for i, a in enumerate(arr):      # (i + 1) // 2 := the number of previous odd-length subarrays.      currEvenSum = prevOddSum + ((i + 1) // 2) * a      # i // 2 + 1 := the number of previous even-length subarrays      # (including 0).      currOddSum = prevEvenSum + (i // 2 + 1) * a      ans += currOddSum      prevEvenSum = currEvenSum      prevOddSum = currOddSum     return ans

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