sockets-Bita

lib/array_intersection.rb

@@ -1,6 +1,62 @@
 # Returns a new array to that contains elements in the intersection of the two input arrays
-# Time complexity: ?
-# Space complexity: ?
+# Time complexity: o(n log m)
+# Space complexity: o(n) n is the number of elements of the smallest array
+# def intersection(array1, array2)
+#   return [] if array1 == [] || array2 == []
+#   return [] if array1 == nil || array2 == nil
+#   intersection = []
+#   array1.each do |i|
+#     array2.each do |j|
+#       if i == j
+#         intersection.push(i)
+#       end
+#     end
+#   end
+#   return intersection
+# end
 def intersection(array1, array2)
-  raise NotImplementedError
+  if array1 == nil || array2 == nil
+    return []
+  else
+    intersection = []
+    if array1.length >= array2.length
+      i = 0
+      array = array1.sort
+      array2.length.times do
+        if binary_search(array, array2[i])
+          intersection.push(array2[i])
+        end
+        i += 1
+      end
+      return intersection
+    else
+      array = array2.sort
+      array1.length.times do
+        if binary_search(array, array1[i])
+          intersection.push(array1[i])
+        end
+        i += 1
+      end
+      return intersection
+    end
+  end
+end
+
+def binary_search(array, value_to_find)
+  mid = array.length / 2
+  i = 0
+  j = array.length - 1
+  while i < j
+    if array[mid] == value_to_find
+      return true
+    elsif array[mid] < value_to_find
+      i = mid + 1
+      j = array.length - 1
+      mid = (i + j) / 2
+    else
+      j = mid - 1
+      mid = (i + j) / 2
+    end
+  end
+  return false
 end