Completed Two-Pointers-1-ContainerWithMostWater

ContainerWithMostWater.java

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+// Time Complexity : O(n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+// Your code here along with comments explaining your approach
+/*
+Maintain 2 pointers one at left and right. We are interested to achieve maximum area which is product of
+length and width.Also, to form a container,between 2 heights, only the lowest length of rod can be considered
+as the rest will overflow.So, we compare and consider the length accordingly. Width can be computed by the
+difference of indices. We move the left pointer if the left rod's height is small, as we want to maximize
+area, if not, we decrement right pointer. This way, we can achieve max area.
+ */
+class Solution {
+    public int maxArea(int[] height) {
+        int left = 0 , right = height.length - 1;
+        int maxArea = 0;
+        while(left < right) {
+            int width = right - left;
+            int length = 0;
+            if(height[left] < height[right]) {
+                length = height[left];
+                left++;
+            }
+            else {
+                length = height[right];
+                right--;
+            }
+            maxArea = Math.max(maxArea, length * width);
+        }
+        return maxArea;
+    }
+}