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| 1 | +# 53. Maximum Subarray |
| 2 | + |
| 3 | +## Kadane's Algothrim |
| 4 | +- Run-time: O(N) |
| 5 | +- Space: O(1) |
| 6 | +- N = Number of nodes in tree |
| 7 | + |
| 8 | +Kadane's Algothrim is a good thing to know. |
| 9 | +There are proofs out there explaining why this works if you are interested. |
| 10 | +Overall its a dynamic programming solution at its core. |
| 11 | + |
| 12 | +Because the question is asking for a contiguous subarray, we can use the previous sums to find our max sum for a given n. |
| 13 | +The main idea is that there are two choices to be made, whether (previous sum + n) or (n) is larger. |
| 14 | + |
| 15 | +``` |
| 16 | +class Solution: |
| 17 | + def maxSubArray(self, nums: List[int]) -> int: |
| 18 | + if len(nums) == 0: |
| 19 | + return 0 |
| 20 | + max_sum, prev_sum = nums[0], nums[0] |
| 21 | + for n in nums[1:]: |
| 22 | + prev_sum = max(n, n+prev_sum) |
| 23 | + max_sum = max(max_sum, prev_sum) |
| 24 | + return max_sum |
| 25 | +``` |
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