added increasingTripletSubsequence

README.md

@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|334|[Increasing Triplet Subsequence](https://leetcode.com/problems/increasing-triplet-subsequence/) | [C++](./algorithms/cpp/increasingTripletSubsequence/increasingTripletSubsequence.cpp)|Medium|
 |329|[Longest Increasing Path in a Matrix](https://leetcode.com/problems/longest-increasing-path-in-a-matrix/) | [C++](./algorithms/cpp/longestIncreasingPathInAMatrix/LongestIncreasingPathInAMatrix.cpp)|Medium|
 |328|[Odd Even Linked List](https://leetcode.com/problems/odd-even-linked-list/) | [C++](./algorithms/cpp/oddEvenLinkedList/OddEvenLinkedList.cpp)|Easy|
 |327|[Count of Range Sum](https://leetcode.com/problems/count-of-range-sum/) | [C++](./algorithms/cpp/countOfRangeSum/CountOfRangeSum.cpp)|Hard|

algorithms/cpp/increasingTripletSubsequence/increasingTripletSubsequence.cpp

@@ -0,0 +1,43 @@
+// Source : https://leetcode.com/problems/increasing-triplet-subsequence/
+// Author : Calinescu Valentin
+// Date   : 2016-02-27
+
+/*************************************************************************************** 
+ *
+ * Given an unsorted array return whether an increasing subsequence of length 3 exists
+ * or not in the array.
+ * 
+ * Formally the function should:
+ * Return true if there exists i, j, k 
+ * such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.
+ * Your algorithm should run in O(n) time complexity and O(1) space complexity.
+ * 
+ * Examples:
+ * Given [1, 2, 3, 4, 5],
+ * return true.
+ * 
+ * Given [5, 4, 3, 2, 1],
+ * return false.
+ * 
+ ***************************************************************************************/
+class Solution {
+public:
+    bool increasingTriplet(vector<int>& nums) {
+        bool solution = false;
+        if(nums.size())
+        {
+            int first = nums[0];
+            int second = 0x7fffffff; //MAX_INT so we can always find something smaller than it
+            for(int i = 1; i < nums.size() && !solution; i++)
+            {
+                if(nums[i] > second)
+                    solution = true;
+                else if(nums[i] > first && nums[i] < second)
+                    second = nums[i];
+                else if(nums[i] < first)
+                    first = nums[i];
+            }
+        }
+        return solution;
+    }
+};