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Problem solution · Python

Maximum Number of Non-Overlapping Subarrays With Sum Equals Target

Maximum Number of Non-Overlapping Subarrays With Sum Equals Target: 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
21 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 Number of Non-Overlapping Subarrays With Sum Equals Target, 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

  • 21 lines of Python from the credited upstream file 1546.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 codeMaximum Number of Non-Overlapping Subarrays With Sum Equals Target · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def maxNonOverlapping(self, nums: list[int], target: int) -> int:    # Ending the subarray ASAP always has a better result.    ans = 0    prefix = 0    prefixes = {0}     # Greedily find the subarrays that equal to the target.    for num in nums:      # Check if there is a subarray ends in here and equals to the target.      prefix += num      if prefix - target in prefixes:        # Find one and discard all the prefixes that have been used.        ans += 1        prefix = 0        prefixes = {0}      else:        prefixes.add(prefix)     return ans

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