Add solution for challenge #48

Author: Mauricio Klein

README.md

@@ -47,3 +47,4 @@
 - [x] [Challenge 45: Occurrences of a number in a sorted array](challenge-45/)
 - [x] [Challenge 46: Sudoku solver](challenge-46/)
 - [x] [Challenge 47: Closest coin](challenge-47/)
+- [x] [Challenge 48: Median of window](challenge-48/)

challenge-48/Makefile

@@ -0,0 +1,4 @@
+test:
+	python test_*.py
+
+.PHONY: test

challenge-48/README.md

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+# Median of window
+
+This problem was asked by Microsoft.
+
+## Description
+
+Given an array of numbers "arr" and a window of size "k", return an array with the median of each window of size "k" starting from the left and moving right by one position each time.
+
+Recall that the median of an even-sized list is the average of the two middle numbers.
+
+## Example
+```
+Input:
+  arr: [-1, 5, 13, 8, 2, 3, 3, 1]
+
+Output:
+  [5, 8, 8, 3, 3, 3]
+  # 5 <- median of [-1, 5, 13]
+  # 8 <- median of [5, 13, 8]
+  # 8 <- median of [13, 8, 2]
+  # 3 <- median of [8, 2, 3]
+  # 3 <- median of [2, 3, 3]
+  # 3 <- median of [3, 3, 1]
+```

challenge-48/__init__.py

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+import bisect
+
+#
+# Brute force solution:
+# For each window with "k" elements, sort the window and return the median.
+#   Time  complexity: O(n*klog(k))
+#   Space complexity: O(n) (for the output array)
+#
+
+#
+# Optimized solution:
+# Initialize the window with the "k" first elements using bisect insert to place the element while keeping the
+# window sorted. For every new window, remove the oldest element and add the new one, using bisect insert.
+# Return the median of each window as usual
+#   Time  complexity: O(n*k)
+#   Space complexity: O(n) (for the output array)
+
+#
+# Time  complexity: O(n*k) (n = number of elements in the array)
+# Space complexity: O(n)   (for the output buffer)
+#
+def subarr_mean(arr, k):
+  # Populate the buffer with the first "k" elements of the array,
+  # adding them in order, to keep the array sorted: O(k)
+  buffer = []
+  for i in range(0, k):
+    bisect.insort(buffer, arr[i])
+
+  oldest = arr[0]
+  output = [median(buffer, k)]
+
+  for i in range(k, len(arr)):
+    # Remove the oldest value: O(k)
+    buffer.remove(oldest)
+
+    # Insert the new value on the right place, in
+    # order to keep the buffer sorted: O(k)
+    bisect.insort(buffer, arr[i])
+
+    # Add the median to the output array: O(1)
+    output.append(median(buffer, k))
+
+    # Update the oldest value: O(1)
+    oldest = arr[i - k + 1]
+
+  return output
+
+#
+# Time  complexity: O(1)
+# Space complexity: O(1)
+#
+def median(subarr, k):
+  """
+  Return the median of the subarray based on the value ok "k"
+  """
+  even_k = (k % 2 == 0)
+
+  if even_k:
+    right  = k / 2
+    left = right - 1
+
+    return (subarr[left] + subarr[right]) / 2
+  else:
+    id = k // 2
+
+    return subarr[id]
+

challenge-48/solver.py

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+import unittest
+from solver import subarr_mean 
+
+class TestSolver(unittest.TestCase):
+  def test_closest_coin(self):
+    self.assertEqual(subarr_mean([-1, 5, 13, 8, 2, 3, 3, 1], k=3), [5, 8, 8, 3, 3, 3])
+    self.assertEqual(subarr_mean([-1, 5, 13, 8, 2, 3, 3, 1], k=4), [6, 6, 5, 3, 2   ])
+    self.assertEqual(subarr_mean([1, 2, 3, 4              ], k=1), [1, 2, 3, 4      ])
+    self.assertEqual(subarr_mean([1, 50, 10, 20           ], k=4), [15              ])
+
+if __name__ == "__main__":
+    unittest.main()