Add a solution to the Employee Free Time

Sort/EmployeeFreeTime.swift

@@ -0,0 +1,77 @@
+/**
+ * Question Link: https://leetcode.com/problems/employee-free-time/
+ * Primary idea: Combine and merge sorted arrays. Then iterate through it to get offset between every two elements.
+ * Time Complexity: O(n), Space Complexity: O(n)
+ *
+ * Definition for an Interval.
+ * public class Interval {
+ *     public var start: Int
+ *     public var end: Int
+ *     public init(_ start: Int, _ end: Int) {
+ *         self.start = start
+ *         self.end = end
+ *     }
+ * }
+ */
+
+class EmployeeFreeTime {
+    func employeeFreeTime(_ schedule: [[Interval]]) -> [Interval] {
+        let intervals = mergeIntervals(combineIntervals(schedule))
+        var res = [Interval]()
+
+        for i in 1..<intervals.count {
+            res.append(Interval(intervals[i - 1].end, intervals[i].start))
+        }
+
+        return res
+    }
+
+    private func mergeIntervals(_ intervals: [Interval]) -> [Interval] {
+        var res = [Interval]()
+
+        for interval in intervals {
+            if let last = res.last, last.end >= interval.start {
+                res.removeLast()
+                res.append(Interval(last.start, max(last.end, interval.end))) 
+            } else {
+                res.append(interval)
+            }
+        }
+
+        return res
+    }
+
+    private func combineIntervals(_ schedule: [[Interval]]) -> [Interval] {
+        var res = schedule[0]
+
+        for i in 1..<schedule.count {
+            res = combineTwoIntervals(res, schedule[i])
+        }
+
+        return res
+    }
+
+    private func combineTwoIntervals(_ l: [Interval], _ r: [Interval]) -> [Interval] {
+        var res = [Interval](), i = 0, j = 0
+
+        while i < l.count || j < r.count {
+            if i == l.count {
+                res.append(r[j])
+                j += 1
+            } else if j == r.count {
+                res.append(l[i])
+                i += 1
+            } else {
+                if l[i].start <= r[j].start {
+                    res.append(l[i])
+                    i += 1
+                } else {
+                    res.append(r[j])
+                    j += 1
+                }
+            }
+        }
+
+        return res
+    }
+}