Add count occurrences algorithm

Author: hollance

Count Occurrences/CountOccurrences.playground/Contents.swift

@@ -0,0 +1,67 @@
+//: Playground - noun: a place where people can play
+
+func countOccurrencesOfKey(key: Int, inArray a: [Int]) -> Int {
+  func leftBoundary() -> Int {
+    var low = 0
+    var high = a.count
+    while low < high {
+      let midIndex = low + (high - low)/2
+      if a[midIndex] < key {
+        low = midIndex + 1
+      } else {
+        high = midIndex
+      }
+    }
+    return low
+  }
+  
+  func rightBoundary() -> Int {
+    var low = 0
+    var high = a.count
+    while low < high {
+      let midIndex = low + (high - low)/2
+      if a[midIndex] > key {
+        high = midIndex
+      } else {
+        low = midIndex + 1
+      }
+    }
+    return low
+  }
+  
+  return rightBoundary() - leftBoundary()
+}
+
+
+// Simple test
+
+let a = [ 0, 1, 1, 3, 3, 3, 3, 6, 8, 10, 11, 11 ]
+countOccurrencesOfKey(3, inArray: a)
+
+
+// Test with arrays of random size and contents (see debug output)
+
+import Foundation
+
+func createArray() -> [Int] {
+  var a = [Int]()
+  for i in 0...5 {
+    if i != 2 {  // don't include the number 2
+      let count = Int(arc4random_uniform(UInt32(6))) + 1
+      for _ in 0..<count {
+        a.append(i)
+      }
+    }
+  }
+  return a.sort(<)
+}
+
+for _ in 0..<10 {
+  let a = createArray()
+  print(a)
+
+  // Note: we also test -1 and 6 to check the edge cases.
+  for k in -1...6 {
+    print("\t\(k): \(countOccurrencesOfKey(k, inArray: a))")
+  }
+}

Count Occurrences/CountOccurrences.playground/contents.xcplayground

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+<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
+<playground version='5.0' target-platform='osx'>
+    <timeline fileName='timeline.xctimeline'/>
+</playground>

Count Occurrences/CountOccurrences.playground/timeline.xctimeline

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+<?xml version="1.0" encoding="UTF-8"?>
+<Timeline
+   version = "3.0">
+   <TimelineItems>
+   </TimelineItems>
+</Timeline>

Count Occurrences/CountOccurrences.swift

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+/*
+  Counts the number of times a value appears in an array in O(lg n) time.
+  The array must be sorted from low to high.
+*/
+func countOccurrencesOfKey(key: Int, inArray a: [Int]) -> Int {
+  func leftBoundary() -> Int {
+    var low = 0
+    var high = a.count
+    while low < high {
+      let midIndex = low + (high - low)/2
+      if a[midIndex] < key {
+        low = midIndex + 1
+      } else {
+        high = midIndex
+      }
+    }
+    return low
+  }
+
+  func rightBoundary() -> Int {
+    var low = 0
+    var high = a.count
+    while low < high {
+      let midIndex = low + (high - low)/2
+      if a[midIndex] > key {
+        high = midIndex
+      } else {
+        low = midIndex + 1
+      }
+    }
+    return low
+  }
+
+  return rightBoundary() - leftBoundary()
+}

Count Occurrences/README.markdown

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+# Count Occurrences
+
+Goal: Count how often a certain value appears in an array.
+
+The obvious way to do this is with a [linear search](../Linear Search/) from the beginning of the array until the end, keeping count of how often you come across the value. This is an **O(n)** algorithm.
+
+However, if the array is sorted you can do it much faster, in **O(log n)** time, by using a modification of [binary search](../Binary Search/).
+
+Let's say we have the following array:
+
+	[ 0, 1, 1, 3, 3, 3, 3, 6, 8, 10, 11, 11 ]
+
+If we want to know how often the value `3` occurs, we can do a binary search for `3`. That could give us any of these four indices:
+
+	[ 0, 1, 1, 3, 3, 3, 3, 6, 8, 10, 11, 11 ]
+	           *  *  *  *
+
+But that still doesn't tell you how many other `3`s there are. To find those other `3`s, you'd still have to do a linear search to the left and a linear search to the right. That will be fast enough in most cases, but in the worst case -- when the array consists of nothing but `3`s -- it still takes **O(n)** time.
+
+The trick is to use two binary searches, one to find where the `3`s start (the left boundary), and one to find where they end (the right boundary).
+
+In code this looks as follows:
+
+```swift
+func countOccurrencesOfKey(key: Int, inArray a: [Int]) -> Int {
+  func leftBoundary() -> Int {
+    var low = 0
+    var high = a.count
+    while low < high {
+      let midIndex = low + (high - low)/2
+      if a[midIndex] < key {
+        low = midIndex + 1
+      } else {
+        high = midIndex
+      }
+    }
+    return low
+  }
+  
+  func rightBoundary() -> Int {
+    var low = 0
+    var high = a.count
+    while low < high {
+      let midIndex = low + (high - low)/2
+      if a[midIndex] > key {
+        high = midIndex
+      } else {
+        low = midIndex + 1
+      }
+    }
+    return low
+  }
+  
+  return rightBoundary() - leftBoundary()
+}
+```
+
+Notice that the helper functions `leftBoundary()` and `rightBoundary()` are very similar to the binary search algorithm. The big difference is that they don't stop when they find the search key, but keep going.
+
+To test this algorithm, copy the code to a playground and then do:
+
+```swift
+let a = [ 0, 1, 1, 3, 3, 3, 3, 6, 8, 10, 11, 11 ]
+
+countOccurrencesOfKey(3, inArray: a)  // returns 4
+```
+
+Remember: If you use your own array, make sure it is sorted first!
+
+Let's walk through the example. The array is:
+
+	[ 0, 1, 1, 3, 3, 3, 3, 6, 8, 10, 11, 11 ]
+
+To find the left boundary, we start with `low = 0` and `high = 12`. The first mid index is `6`:
+
+	[ 0, 1, 1, 3, 3, 3, 3, 6, 8, 10, 11, 11 ]
+	                    *
+
+With a regular binary search you'd be done now, but here we're not just looking whether the value `3` occurs or not -- instead, we want to find where it occurs *first*.
+
+Since this algorithm follows the same principle as binary search, we now ignore the right half of the array and calculate the new mid index:
+
+	[ 0, 1, 1, 3, 3, 3 | x, x, x, x, x, x ]
+	           *
+
+Again, we've landed on a `3`, and it's the very first one. But the algorithm doesn't know that, so we split the array again:
+
+	[ 0, 1, 1 | x, x, x | x, x, x, x, x, x ]
+	     *
+
+Still not done. Split again, but this time use the right half:
+
+	[ x, x | 1 | x, x, x | x, x, x, x, x, x ]
+	         *
+
+The array cannot be split up any further, which means we've found the left boundary, at index 3.
+
+Now let's start over and try to find the right boundary. This is very similar, so I'll just show you the different steps:
+
+	[ 0, 1, 1, 3, 3, 3, 3, 6, 8, 10, 11, 11 ]
+	                    *
+
+	[ x, x, x, x, x, x, x | 6, 8, 10, 11, 11 ]
+	                              *
+
+	[ x, x, x, x, x, x, x | 6, 8, | x, x, x ]
+	                           *
+
+	[ x, x, x, x, x, x, x | 6 | x | x, x, x ]
+	                        *
+
+The right boundary is at index 7. The difference between the two boundaries is 7 - 3 = 4, so the number `3` occurs four times in this array.
+
+Each binary search took 4 steps, so in total this algorithm took 8 steps. Not a big gain on an array of only 12 items, but the bigger the array, the more efficient this algorithm becomes. For a sorted array with 1,000,000 items, it only takes 2x20 = 40 steps to count the number of occurrences for any particular value.
+
+By the way, if the value you're looking for is not in the array, then `rightBoundary()` and `leftBoundary()` return the same value and so the difference between them is 0.
+
+This is an example of how you can modify the basic binary search to solve other algorithmic problems as well. Of course, it does require that the array is sorted.
+
+*Written by Matthijs Hollemans*

README.markdown

@@ -28,9 +28,9 @@ If you're new to algorithms and data structures, here are a few good ones to sta
 
 ### Searching
 
-- [Linear Search](Linear Search/)
+- [Linear Search](Linear Search/). Find an element in an array.
 - [Binary Search](Binary Search/). Quickly find elements in a sorted array.
-- Count Occurrences
+- [Count Occurrences](Count Occurrences/). Count how often a value appears in an array.
 - Select Minimum / Maximum
 - Select k-th Largest Element
 - Selection Sampling