Added Solution for LeetCode:11-'Container With Most Water', under Two Pointers

src/TwoPointers/ContainerWithMostWater_11/Main.java

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+package TwoPointers.ContainerWithMostWater_11;
+
+public class Main {
+    public static void main(String[] args) {
+        Solution s = new Solution();
+
+        int[] input = {1,8,6,2,5,4,8,3,7};
+        System.out.println(s.getContainerWithMostWater(input));
+    }
+}

src/TwoPointers/ContainerWithMostWater_11/Solution.java

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+package TwoPointers.ContainerWithMostWater_11;
+
+import java.lang.Math;
+
+/**
+ * 11. Container With Most Water
+ * 
+ * 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.
+ * 
+ *  Input: heights = [1,8,6,2,5,4,8,3,7]
+    Output: 49
+    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.
+
+    leetcode link : https://leetcode.com/problems/container-with-most-water/
+ *  */
+
+class Solution {
+  public int getContainerWithMostWater(int[] heights) {
+    int l = 0;
+    int r = heights.length - 1;
+    int maxArea = 0;
+
+    while (l < r) {
+      int min = Math.min(heights[l], heights[r]);
+      int curr = min * (r - l);
+      maxArea = Math.max(maxArea, curr);
+
+      if (heights[r] == heights[l]) {
+        l++;
+        r--;
+      }
+      else if (heights[l] < heights[r]) l++;
+      else r--;
+    }
+    return maxArea;
+  }
+}