Merge pull request kodecocodes#634 from nb2998/nb2998-bigO

Updated Big-O Notation.markdown

Author: kelvinlauKL

Big-O Notation.markdown

@@ -15,8 +15,130 @@ Big-O | Name | Description
 **O(n^2)** | quadratic | **Kinda slow.** If you have 100 items, this does 100^2 = 10,000 units of work. Doubling the number of items makes it four times slower (because 2 squared equals 4). Example: algorithms using nested loops, such as insertion sort.
 **O(n^3)** | cubic | **Poor performance.** If you have 100 items, this does 100^3 = 1,000,000 units of work. Doubling the input size makes it eight times slower. Example: matrix multiplication.
 **O(2^n)** | exponential | **Very poor performance.** You want to avoid these kinds of algorithms, but sometimes you have no choice. Adding just one bit to the input doubles the running time. Example: traveling salesperson problem.
-**O(n!)** | factorial | **Intolerably slow.** It literally takes a million years to do anything.
+**O(n!)** | factorial | **Intolerably slow.** It literally takes a million years to do anything.  
+  
 
+Below are some examples for each category of performance:
+
+**O(1)**
+
+  The most common example with O(1) complexity is accessing an array index.
+  
+  ```swift
+  let value = array[5]
+  ```
+    
+  Another example of O(1) is pushing and popping from Stack.
+  
+ 
+**O(log n)**
+
+  ```swift
+  var j = 1
+  while j < n {
+    // do constant time stuff
+    j *= 2
+  }
+  ```  
+  
+  Instead of simply incrementing, 'j' is increased by 2 times itself in each run.
+  
+  Binary Search Algorithm is an example of O(log n) complexity.
+  
+  
+**O(n)**
+
+  ```swift
+  for i in stride(from: 0, to: n, by: 1) {
+    print(array[i])
+  }
+  ```
+  
+  Array Traversal and Linear Search are examples of O(n) complexity.  
+  
+  
+**O(n log n)**
+
+  ```swift
+  for i in stride(from: 0, to: n, by: 1) {
+  var j = 1
+    while j < n {
+      j *= 2
+      // do constant time stuff
+    }
+  }
+  ```
+  
+  OR
+  
+  ```swift
+  for i in stride(from: 0, to: n, by: 1) {
+    func index(after i: Int) -> Int? { // multiplies `i` by 2 until `i` >= `n`
+      return i < n ? i * 2 : nil 
+    }
+    for j in sequence(first: 1, next: index(after:)) {
+      // do constant time stuff
+    }
+  }
+  ```
+  
+  Merge Sort and Heap Sort are examples of O(n log n) complexity.  
+  
+  
+**O(n^2)**
+
+  ```swift
+  for i  in stride(from: 0, to: n, by: 1) {
+    for j in stride(from: 1, to: n, by: 1) {
+      // do constant time stuff
+    }
+  }
+  ```
+  
+  Traversing a simple 2-D array and Bubble Sort are examples of O(n^2) complexity.
+  
+  
+**O(n^3)**
+
+  ```swift
+  for i in stride(from: 0, to: n, by: 1) {
+    for j in stride(from: 1, to: n, by: 1) {
+      for k in stride(from: 1, to: n, by: 1) {
+        // do constant time stuff
+      }
+    }
+  }
+  ```  
+  
+**O(2^n)**
+
+  Algorithms with running time O(2^N) are often recursive algorithms that solve a problem of size N by recursively solving two smaller problems of size N-1.
+  The following example prints all the moves necessary to solve the famous "Towers of Hanoi" problem for N disks.
+
+  ```swift
+  func solveHanoi(N: Int, from: String, to: String, spare: String) {
+    guard n >= 1 else { return }
+    if N > 1 {
+      solveHanoi(N: N - 1, from: from, to: spare, spare: to)
+    } else {
+      solveHanoi(N: N-1, from: spare, to: to, spare: from)
+    }
+  }
+  ```
+  
+  
+**O(n!)**
+
+  The most trivial example of function that takes O(n!) time is given below.
+
+  ```swift
+  func nFacFunc(n: Int) {
+    for i in stride(from: 0, to: n, by: 1) {
+      nFactFunc(n - 1)
+    }
+  }
+  ``` 
+  
 Often you don't need math to figure out what the Big-O of an algorithm is but you can simply use your intuition. If your code uses a single loop that looks at all **n** elements of your input, the algorithm is **O(n)**. If the code has two nested loops, it is **O(n^2)**. Three nested loops gives **O(n^3)**, and so on.
 
 Note that Big-O notation is an estimate and is only really useful for large values of **n**. For example, the worst-case running time for the [insertion sort](Insertion%20Sort/) algorithm is **O(n^2)**. In theory that is worse than the running time for [merge sort](Merge%20Sort/), which is **O(n log n)**. But for small amounts of data, insertion sort is actually faster, especially if the array is partially sorted already!