Algorithm Notes
Summary: Maximum Subarray — notes not yet curated.
Time: Estimate via loops/recurrences; common classes: O(1), O(log n), O(n), O(n log n), O(n^2)
Space: Count auxiliary structures and recursion depth.
Tip: See the Big-O Guide for how to derive bounds and compare trade-offs.
Big-O Guide
Source
"""
Maximum Subarray
TODO: Add problem description
"""
from src.interview_workbook.leetcode._registry import register_problem
from src.interview_workbook.leetcode._types import Category, Difficulty
class Solution:
def solve(self, nums):
"""Kadane's algorithm to find maximum subarray sum."""
max_sum = nums[0]
current_sum = nums[0]
for n in nums[1:]:
current_sum = max(n, current_sum + n)
max_sum = max(max_sum, current_sum)
return max_sum
def demo() -> str:
"""Run a demo for the Maximum Subarray problem."""
nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]
print(f"Input nums: {nums}")
s = Solution()
result = s.solve(nums)
print(f"Final result: {result}")
return f"Maximum Subarray result for {nums} -> {result}"
if __name__ == "__main__":
demo()
register_problem(
id=53,
slug="maximum_subarray",
title="Maximum Subarray",
category=Category.GREEDY,
difficulty=Difficulty.MEDIUM,
tags=["array", "divide_and_conquer", "dynamic_programming"],
url="https://leetcode.com/problems/maximum-subarray/",
notes="",
)