@@ -28,6 +28,7 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Maximum Swap](problems/src/array/MaximumSwap.java) (Medium)
 - [H-Index](problems/src/array/HIndex.java) (Medium)
 - [Insert Interval](problems/src/array/InsertInterval.java) (Hard)
+- [Increasing Triplet Subsequence](problems/src/array/IncreasingTripletSubsequence.java) (Medium)
 
 #### [Backtracking](problems/src/backtracking)
 
 
@@ -0,0 +1,48 @@
+package array;
+
+import java.util.Arrays;
+
+/**
+ * Created by gouthamvidyapradhan on 17/12/2017.
+ * 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.
+ */
+public class IncreasingTripletSubsequence {
+
+    /**
+     * Main method
+     * @param args
+     * @throws Exception
+     */
+    public static void main(String[] args) throws Exception{
+        int[] A = {1, 2, 3, 4, 5};
+        System.out.println(new IncreasingTripletSubsequence().increasingTriplet(A));
+    }
+
+    public boolean increasingTriplet(int[] nums) {
+        int[] A = new int[3];
+        Arrays.fill(A, Integer.MAX_VALUE);
+        for (int num : nums) {
+            if (num < A[0]) {
+                A[0] = num;
+            } else if (num < A[1] && num > A[0]) {
+                A[1] = num;
+            } else if (num < A[2] && num > A[1]) {
+                return true;
+            }
+        }
+        return false;
+    }
+}
+