3Sum and 4Sum

Follow up 2Sum problem. How to solve 3Sum and 4Sum problem? What’s their relationships? Could we get some generic idea to solve this kind of problems? I give the explanation here.

Author: remlostime

3Sum and 4Sum/3Sum.playground/Contents.swift

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+func ThreeSum(_ nums: [Int], targetSum: Int) -> [[Int]] {
+    var a = nums.sorted()
+    var ret: [[Int]] = []
+    
+    for i in 0..<a.count {
+        if i != 0 && a[i] == a[i-1] {
+            continue
+        }
+        
+        var j = i + 1
+        var k = nums.count - 1
+        
+        while j < k {
+            let sum = a[i] + a[j] + a[k]
+            
+            if sum == targetSum {
+                ret.append([a[i], a[j], a[k]])
+                j += 1
+            } else if sum < targetSum {
+                j += 1
+            } else {
+                k -= 1
+            }
+            
+            if sum == targetSum {
+                while j < k {
+                    var flag = false
+                    if j != 0 && a[j] == a[j-1] {
+                        j += 1
+                        flag = true
+                    }
+                    
+                    if (k != a.count - 1) && a[k] == a[k+1] {
+                        k -= 1
+                        flag = true
+                    }
+                    
+                    if (!flag) {
+                        break
+                    }
+                }
+            }
+        }
+    }
+    
+    return ret
+}
+
+// Answer: [[-1, 0, 1], [-1, -1, 2]]
+ThreeSum([-1, 0, 1, 2, -1, -4], targetSum: 0)
+
+// Answer: [[-1, -1, 1]]
+ThreeSum([-1, 1, 1, -1], targetSum: 1)

3Sum and 4Sum/3Sum.playground/contents.xcplayground

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

3Sum and 4Sum/4Sum.playground/Contents.swift

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+func FourSum(_ nums: [Int], target: Int) -> [[Int]] {
+    let a = nums.sorted()
+    var ret: [[Int]] = []
+    
+    for i in 0..<nums.count {
+        if i != 0 && a[i] == a[i-1] {
+            continue
+        }
+        for j in i+1..<nums.count {
+            if j != i+1 && a[j] == a[j-1] {
+                continue
+            }
+            
+            var k = j + 1
+            var t = nums.count - 1
+            
+            while k < t {
+                let sum = a[i] + a[j] + a[k] + a[t]
+                
+                if sum == target {
+                    ret.append([a[i], a[j], a[k], a[t]])
+                    k += 1
+                } else if sum < target {
+                    k += 1
+                } else {
+                    t -= 1
+                }
+                
+                if sum == target {
+                    while k < t {
+                        var flag = true
+                        if k != j + 1 && a[k] == a[k-1] {
+                            k += 1
+                            flag = false
+                        }
+                        
+                        if t != nums.count - 1 && a[t] == a[t+1] {
+                            t -= 1
+                            flag = false
+                        }
+                        
+                        if flag {
+                            break
+                        }
+                    }
+                }
+            }
+        }
+    }
+    
+    return ret
+}
+
+/* Answer
+[
+    [-1,  0, 0, 1],
+    [-2, -1, 1, 2],
+    [-2,  0, 0, 2]
+]
+*/
+FourSum([1, 0, -1, 0, -2, 2], target: 0)

3Sum and 4Sum/4Sum.playground/contents.xcplayground

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

3Sum and 4Sum/README.md

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+# 3Sum and 4Sum
+
+## 3Sum
+Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? 
+Find all unique triplets in the array which gives the sum of zero.
+
+**Note**: The solution set must not contain duplicate triplets.
+
+> For example, given array S = [-1, 0, 1, 2, -1, -4](),
+> A solution set is:
+> [
+> ]()>  [-1, 0, 1](),
+>  [-1, -1, 2]()
+> ]
+
+### Solution
+Like [2Sum][5], we can sort the array first, then use 2 pointers method to solve the problem. But here, we need 3 numbers, so we need add one more pointer. But how to play with 3 pointers?
+
+The key idea here is we split the array into 2 parts, if we have a pointer `i`, let’s assume we got `i` position already. What’ next? Actually, next is just a 2Sum problem right? We can just apply 2 pointers method in 2Sum. But what’s 2Sum’s target sum? Is it `target2Sum = target3Sum - nums[i]` ? That’s awesome! 
+
+How we figure out `i`? We can loop i from `0` to `nums.count - 1`.
+
+How we avoid duplicate triplets? Since the array has been sorted, we just need to check if `a[i] == a[i-1]`. Why? Because if `a[i] == a[i-1]`, it means that `a[i-1]` already covers all `a[i]` solutions, we should not do it, otherwise it will have duplicate answers. The same for `j` and `k` which runs for 2Sum.
+
+So, we successfully downgrade 3Sum to 2Sum problem with some tricks. Next, you will see how we downgrade 4Sum to 3Sum.
+
+## 4Sum
+Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
+
+**Note**: The solution set must not contain duplicate quadruplets.
+
+### Solution
+After 3Sum, you should have feeling actually we just need a same idea to downgrade it to 3Sum, and then 2Sum, and then solve it.
+
+How? I will leave it as a challenge for you to figure out first and see if you really master the idea behind this kind of problems.
+
+Feel free to check out the solution if you are blocked.
+
+## Where to go next?
+If it’s a KSum, and `K` is a big number, do we need to create `K` pointers and solve it?
+
+I will write another topic to present how we will solve this KSum problem with a generic way soon.
+
+
+
+
+
+[5]:	https://github.com/raywenderlich/swift-algorithm-club/tree/master/Two-Sum%20Problem