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Author: 17tanya

TwoPointer/ContainerWithMostWater.java

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+//LeetCode 11. Container with most water
+//Question - https://leetcode.com/problems/container-with-most-water/
+
+class Solution {
+    public int maxArea(int[] height) {
+        int start = 0;
+        int end = height.length - 1;
+        int maxWater = 0;
+        
+        /*
+        Brute force solution would be to evaluate all containers. This will take
+        O(n^2) time. This can be avoided by considering the fact that the 
+        maximum water a container can hold is restricted by the minimum of the
+        two sides. The minimum value acts as a limiting factor. 
+        
+        There is no point in pairing-up this minimum value with other greater values
+        as the height is limited by this value.
+        
+        To gain maximum profit from this value, we need to pair it with the
+        farthest value which is equal to or greater than this value. This is 
+        the maximum water we can retain with this value and so it is safe to
+        move the pointer pointing at this minimum value.
+        */
+        
+        while(start < end){
+            int water = Math.min(height[start], height[end]) * (end - start);
+            maxWater = Math.max(maxWater, water);
+            
+            if(height[start] < height[end]) start++;
+            else end--;
+        }
+        
+        return maxWater;
+    }
+}