Updated for FourSum.

Author: kelvinlauKL

3Sum and 4Sum/README.md

@@ -78,11 +78,11 @@ func threeSum<T: BidirectionalCollection>(_ collection: T, target: T.Element) ->
   var ret: [[T.Element]] = []
   var l = sorted.startIndex
   
-  while l < sorted.endIndex { defer { sorted.formUniqueIndex(after: &l) }
+  ThreeSum: while l < sorted.endIndex { defer { sorted.formUniqueIndex(after: &l) }
     var m = sorted.index(after: l)
     var r = sorted.index(before: sorted.endIndex)
     
-    while m < r && r < sorted.endIndex { 
+    TwoSum: while m < r && r < sorted.endIndex { 
       let sum = sorted[l] + sorted[m] + sorted[r]
       if sum == target {
         ret.append([sorted[l], sorted[m], sorted[r]])
@@ -101,13 +101,60 @@ func threeSum<T: BidirectionalCollection>(_ collection: T, target: T.Element) ->
 ```
 
 ## 4Sum
+
 Given an array S of n integers, find all subsets of the array with 4 values where the 4 values sum up to a target number. 
 
 **Note**: The solution set must not contain duplicate quadruplets.
 
 ### Solution
-After 3Sum, we already have the idea to change to a problem to a familiar problem we solved before. So, the idea here is straightforward. We just need to downgrade 4Sum to 3Sum. Then we can solve 4Sum.
 
-It's easy to think that we loop the array and get the first the element, then the rest array is 3Sum problem. Since the code is pretty simple, I will avoid duplicate introducation here.
+Foursum is a very straightforward extension to the threeSum algorithm. In threeSum, you kept track of 3 indices:
+
+```
+      m ->      <- r
+[-4, -1, -1, 0, 1, 2]
+  l   
+```
+
+For fourSum, you'll keep track of 4:
+
+```
+         mr ->  <- r
+[-4, -1, -1, 0, 1, 2]
+  l  ml -> 
+```
+
+Here's the code for it (notice it is very similar to 3Sum):
+
+```
+func fourSum<T: BidirectionalCollection>(_ collection: T, target: T.Element) -> [[T.Element]] where T.Element: Numeric & Comparable {
+  let sorted = collection.sorted()
+  var ret: [[T.Element]] = []
+  
+  var l = sorted.startIndex
+  FourSum: while l < sorted.endIndex { defer { sorted.formUniqueIndex(after: &l) }
+    var ml = sorted.index(after: l)
+    
+    ThreeSum: while ml < sorted.endIndex { defer { sorted.formUniqueIndex(after: &ml) }
+      var mr = sorted.index(after: ml)
+      var r = sorted.index(before: sorted.endIndex)
+      
+      TwoSum: while mr < r && r < sorted.endIndex {
+        let sum = sorted[l] + sorted[ml] + sorted[mr] + sorted[r]
+        if sum == target {
+          ret.append([sorted[l], sorted[ml], sorted[mr], sorted[r]])
+          sorted.formUniqueIndex(after: &mr)
+          sorted.formUniqueIndex(before: &r)
+        } else if sum < target {
+          sorted.formUniqueIndex(after: &mr)
+        } else {
+          sorted.formUniqueIndex(before: &r)
+        }
+      }
+    }
+  }
+  return ret
+}
+```
 
 [5]:	https://github.com/raywenderlich/swift-algorithm-club/tree/master/Two-Sum%20Problem