add py3 soln max subarray

0643_maximum-average-subarray-i/README.md

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+You are given an integer array `nums` consisting of `n` elements, and an integer `k`.
+
+Find a contiguous subarray whose **length is equal to** `k` that has the maximum average value and return _this value_. Any answer with a calculation error less than `10-5` will be accepted.
+
+**Example 1:**
+
+    Input: nums = [1,12,-5,-6,50,3], k = 4
+    Output: 12.75000
+    Explanation: Maximum average is (12 - 5 - 6 + 50) / 4 = 51 / 4 = 12.75
+    
+
+**Example 2:**
+
+    Input: nums = [5], k = 1
+    Output: 5.00000
+    
+
+**Constraints:**
+
+*   `n == nums.length`
+*   `1 <= k <= n <= 105`
+*   `-104 <= nums[i] <= 104`

0643_maximum-average-subarray-i/python3/solution.py

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+# Time: O(N)
+# Space: O(1)
+class Solution:
+    def findMaxAverage(self, nums: List[int], k: int) -> float:
+        # Sliding window approach
+        max_average = float("-inf")
+        moving_sum = 0.0
+        window_start = 0
+        
+        for i in range(0, len(nums)):
+            moving_sum += nums[i]
+            
+            # Need to start computing averages only at this point.
+            # e.g. Suppose nums = [0, 1, 2, 3] and k = 3
+            if i >= k - 1:
+                # e.g. When i = 3 - 1 = 2, start checking for max average
+                max_average = max(moving_sum / k, max_average)
+                
+                # e.g. Remove element at beginning of window, i.e 0 from [0...2]
+                moving_sum -= nums[window_start]
+                
+                # e.g. Move window position from 0 to 1
+                window_start += 1
+        
+        return max_average