diff --git a/Task 1/Container with most water/Question.md b/Task 1/Container with most water/Question.md
index bac07de7..689d67c2 100644
--- a/Task 1/Container with most water/Question.md	
+++ b/Task 1/Container with most water/Question.md	
@@ -1,3 +1,34 @@
-Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of the line i is at (i, ai) and (i, 0). Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.
+Given an array of n non-negative integers a1, a2, ..., an , where each element represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of a line i is at (i, ai) and (i, 0). Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.
 
 Note: You may not slant the container.
+
+Example:<br>
+a) Input: n=9<br>
+lines = [1,8,6,2,5,4,8,3,7]<br>
+Output: 49<br>
+Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water the container can contain is 49 which lies between the index(1,8) i.e, between the line 8 at 1st position and line 7 at 8th position.
+
+
+b) Input: n=5<br>
+lines = [4,3,2,1,4]<br>
+Output: 16<br>
+Explanation: The container containing most water lies between the index 0 and 4 i.e, between the line 4 at 0th position and line 4 at 4th position.
+
+
+Approach: Two Pointer Approach
+
+The max area is calculated by the following formula:
+Area= (j - i) * min(lines[i], lines[j])<br>
+
+We should choose (i, j) so that Area is max. Note that i, j go through the range (1, n) and j > i.
+The simple way is to take all possibilities of (i, j) and compare all obtained Area. The time complexity will be O(n)^2.
+
+But, What we gonna do is to choose all possibilities of (i, j) in a wise way. If:
+lines[i] < lines[j] we will check the next (i+1, j) (or move i to the right)
+lines[i] >= lines[j] we will check the next (i, j-1) (or move j to the left)
+Here is the explaination for that:
+When lines[i] < lines[j] , we don't need to calculate all (i, j-1), (i, j-2), .... Why? because these max areas will be smaller than our Area at (i, j).
+
+NOTE: The Time complexity of this solution is O(n).
+
+
diff --git a/Task 1/Next_Greater_Element_II/NextGreaterSet.java b/Task 1/Next_Greater_Element_II/NextGreaterSet.java
new file mode 100644
index 00000000..a5d5f67d
--- /dev/null
+++ b/Task 1/Next_Greater_Element_II/NextGreaterSet.java	
@@ -0,0 +1,69 @@
+import java.util.Arrays;
+
+public class NextGreaterSet
+{
+	// Utility function to swap two digit
+	static void swap(char ar[], int i, int j)
+	{
+		char temp = ar[i];
+		ar[i] = ar[j];
+		ar[j] = temp;
+	}
+	static void findNext(char ar[], int n)
+	{
+		int i;
+		
+		// I) Start from the right most digit
+		// and find the first digit that is smaller
+		// than the digit next to it.
+		for (i = n - 1; i > 0; i--)
+		{
+			if (ar[i] > ar[i - 1]) {
+				break;
+			}
+		}
+		
+		// If no such digit is found, then all
+		// digits are in descending order means
+		// there cannot be a greater number with
+		// same set of digits
+		if (i == 0)
+		{
+			System.out.println("Not possible");
+		}
+		else
+		{
+			int x = ar[i - 1], min = i;
+			
+			// II) Find the smallest digit on right
+			// side of (i-1)'th digit that is greater
+			// than number[i-1]
+			for (int j = i + 1; j < n; j++)
+			{
+				if (ar[j] > x && ar[j] < ar[min])
+				{
+					min = j;
+				}
+			}
+
+			// III) Swap the above found smallest
+			// digit with number[i-1]
+			swap(ar, i - 1, min);
+
+			// IV) Sort the digits after (i-1)
+			// in ascending order
+			Arrays.sort(ar, i, n);
+			System.out.print("Next number with same" +
+									" set of digits is ");
+			for (i = 0; i < n; i++)
+				System.out.print(ar[i]);
+		}
+	}
+
+	public static void main(String[] args)
+	{
+		char digits[] = { '5','3','4','9','7','6' };
+		int n = digits.length;
+		findNext(digits, n);
+	}
+}
diff --git a/Task 1/Next_Greater_Element_II/greater_number_II.md b/Task 1/Next_Greater_Element_II/greater_number_II.md
new file mode 100644
index 00000000..3fd5f423
--- /dev/null
+++ b/Task 1/Next_Greater_Element_II/greater_number_II.md	
@@ -0,0 +1,16 @@
+Given a number n, find the smallest number that has the same set of digits as n and is greater than n. If n is the greatest possible number with its set of digits, then print “not possible”.
+
+Examples: 
+For simplicity of implementation, we have considered the input number as a string. 
+
+Input:  n = "218765"
+Output: "251678"
+
+Input:  n = "1234"
+Output: "1243"
+
+Input: n = "4321"
+Output: "Not Possible"
+
+Input: n = "534976"
+Output: "536479"
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