Create 056_merge_intervals.md

Author: Joseph Luce

leetcode/medium/056_merge_intervals.md

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+# 56. Merge Intervals
+
+## Solution
+- Runtime: O(Nlog(N))
+- Space: O(1) (Assuming result isn't considered extra space)
+- N = Number of elements in intervals
+
+By sorting the intervals, we can then recreate the result by comparing previous intervals with the current interval from left to right.
+In this way, we can check if the two intervals overlap each other after the sort.
+
+```
+class Solution:
+    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
+        if len(intervals) == 0:
+            return []
+        intervals.sort(key=lambda x: x[0])
+        result = list()
+        new_interval = intervals[0]
+        for interval in intervals[1:]:
+            if interval[0] <= new_interval[1]: # do they overlap?
+                new_interval[1] = max(interval[1], new_interval[1])
+            else:
+                result.append(new_interval)
+                new_interval = interval
+        result.append(new_interval)
+        return result
+```
+
+## Brownie Points
+You can do some more optimizations for some inputs if you mention binary search.
+Since after the sort, we can binary search to find the index of the furthest right interval that is overlapping or itself.