Added 11 Container With Most Water

java/11-Container-With-Most-Water.java

@@ -0,0 +1,32 @@
+class Solution {
+    public int maxArea(int[] height) {
+        int left = 0;
+        int right = height.length - 1;
+        int res = 0;
+        while (left < right) {
+            int containerLength = right - left;
+            int area = containerLength * Math.min(height[left], height[right]);
+            res = Math.max(res, area);
+            if (height[left] < height[right]) {
+                left++;
+            } else {
+                right--;
+            }
+        }
+        return res;
+    }
+}
+
+/*
+ * Using two pointers approach. The container length is the right pointer
+ * subtract by the left pointer.
+ * Then we take the container length and multiply it by the minimum value of the
+ * height at left and the height at right, since the minimum value
+ * determines how much water we can hold. If the height at left is smaller than
+ * right, we move the left for a potential bigger height.
+ * If the left point and the right pointer have equal value, we can move either
+ * of them, it doesn't matter.
+ * 
+ * Time: O(n)
+ * Space: O(1)
+ */