From 319cba8ae3efc7e2e627c16b0ae1c4c81110d149 Mon Sep 17 00:00:00 2001
From: Jidong Xiao <xiaoj8@rpi.edu>
Date: Thu, 14 Sep 2023 14:53:05 -0400
Subject: [PATCH] adding merge interval, non member operator overloading
 solution

---
 leetcode/p56_mergeintervals.cpp | 36 +++++++++++++++++++++++++++++++++
 1 file changed, 36 insertions(+)
 create mode 100644 leetcode/p56_mergeintervals.cpp

diff --git a/leetcode/p56_mergeintervals.cpp b/leetcode/p56_mergeintervals.cpp
new file mode 100644
index 0000000..eae2ca4
--- /dev/null
+++ b/leetcode/p56_mergeintervals.cpp
@@ -0,0 +1,36 @@
+// sort a vector (which only has two elements) based on the first element
+bool operator<(vector<int> &A, vector<int> &B){
+    if(A[0]<B[0]){
+        return true;
+    }else{
+        return false;
+    }
+}
+
+class Solution {
+public:
+    vector<vector<int>> merge(vector<vector<int>>& intervals) {
+        if(intervals.empty()){
+            return {};
+        }
+
+        std::sort(intervals.begin(), intervals.end());
+        vector<vector<int>> result;
+        int size = intervals.size();
+        // traverse the vector
+        for(int i=0;i<size;i++){
+            // we use back() to access a vector's last element
+            // if result is empty,
+            // or if the current result does not have any overlap with the interval we are currently visiting, and push back this currently visited interval
+            if(result.empty() || intervals[i][0] > (result.back())[1]){
+                result.push_back(intervals[i]);
+            }else{
+                // if there is an overlap, then update the last element's second element, if it's smaller than the second element of the currently visited interval.
+                if((result.back())[1] < intervals[i][1]){
+                    (result.back())[1] = intervals[i][1];   
+                }
+            }
+        }
+        return result;
+    }
+};