diff --git a/Medium/56.Merge_Intervals.cpp b/Medium/56.Merge_Intervals.cpp
new file mode 100644
index 0000000..f6b701a
--- /dev/null
+++ b/Medium/56.Merge_Intervals.cpp
@@ -0,0 +1,48 @@
+/*
+Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals,
+ and return an array of the non-overlapping intervals that cover all the intervals in the input.
+
+Example 1:
+Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
+Output: [[1,6],[8,10],[15,18]]
+
+1----3                      => 1----6
+  2----6
+        8----10             => 8----10  
+                15----18    => 15----18
+
+*/
+
+
+class Solution {
+public:
+    vector<vector<int>> merge(vector<vector<int>>& intervals) {
+        
+        vector<vector<int>> ans;
+        sort(intervals.begin(), intervals.end());
+        
+        int n = intervals.size();
+        for(int i=0; i<n; i++)
+        {
+            int s1 = intervals[i][0], e1 = intervals[i][1];
+            for(int j=i+1; j<n; j++)
+            {
+                int s2 = intervals[j][0], e2 = intervals[j][1];
+                // if overlap
+                if(s2 <= e1){
+                    e1 = max(e1, e2);
+                    i = j;
+                }
+                // else break out of inner loop and add to ans
+                else
+                {
+                    i=j-1;
+                    break;
+                }
+            }
+            ans.push_back({s1, e1});
+        }   
+        
+        return ans;
+    }
+};
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