From 859f8161918f09b0bdfb9c36e5db662cd4c2a83c Mon Sep 17 00:00:00 2001
From: Jidong Xiao <xiaoj8@rpi.edu>
Date: Tue, 7 Nov 2023 00:07:43 -0500
Subject: [PATCH] adding unordered set problem, O(n) solution

---
 .../p128_longest_consecutive_sequence.cpp     | 30 +++++++++++++++++++
 1 file changed, 30 insertions(+)
 create mode 100644 leetcode/p128_longest_consecutive_sequence.cpp

diff --git a/leetcode/p128_longest_consecutive_sequence.cpp b/leetcode/p128_longest_consecutive_sequence.cpp
new file mode 100644
index 0000000..1361680
--- /dev/null
+++ b/leetcode/p128_longest_consecutive_sequence.cpp
@@ -0,0 +1,30 @@
+class Solution {
+public:
+    // Question: why is this an O(n) solution when we have a nested loop? Because the inner while loop will only be used if *itr1 is the beginning of the sequence, which means each element will only be visited 2 or 3 times.
+    int longestConsecutive(vector<int>& nums) {
+        int len=0;
+        std::unordered_set<int> set1;
+        int size = nums.size();
+        // store unique elements in nums in set1
+        for(int i=0;i<nums.size();i++){
+            set1.insert(nums[i]);
+        }
+        std::unordered_set<int>::iterator itr1 = set1.begin();
+        while(itr1!=set1.end()){
+            // clearly *itr1 is in the set, because that's the meaning of iteration; and if *itr1-1 is not in the set, then we know *itr1 is the beginning of a sequence.
+            if(!set1.count(*itr1-1)){
+                int x = *itr1+1;
+                // now that *itr1 is the beginning of a sequence, how about *itr1+1?
+                while(set1.count(x)){
+                    x++;
+                }
+		// when we get out this above while loop, it's time to update len, if needed.
+                if(x-*itr1>len){
+                    len = x-*itr1;
+                }
+            }
+            itr1++;
+        }
+        return len;
+    }
+};