Added question 053.

Author: luceCoding

Diff for: leetcode/easy/053_maximum_subarray.md

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+# 53. Maximum Subarray
+
+## Kadane's Algothrim
+- Run-time: O(N)
+- Space: O(1)
+- N = Number of nodes in tree
+
+Kadane's Algothrim is a good thing to know.
+There are proofs out there explaining why this works if you are interested.
+Overall its a dynamic programming solution at its core.
+
+Because the question is asking for a contiguous subarray, we can use the previous sums to find our max sum for a given n.
+The main idea is that there are two choices to be made, whether (previous sum + n) or (n) is larger.
+
+```
+class Solution:
+    def maxSubArray(self, nums: List[int]) -> int:
+        if len(nums) == 0:
+            return 0
+        max_sum, prev_sum = nums[0], nums[0]
+        for n in nums[1:]:
+            prev_sum = max(n, n+prev_sum)
+            max_sum = max(max_sum, prev_sum)
+        return max_sum
+```