# The Problem:

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 line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container.

# Notes:

- Use two pointers, start from beginning and end, each iteration move the index which has smaller value towards middle.
- Only update the max Area when new index has bigger value than previous value, since the distance already decreasing, if the height is also decreasing then can’t get a larger area than previous.
- Mistake I made: Be careful in the inner while loop condition, still need to always hold start<end.

# Java Solution:

public class Solution { public int maxArea(int[] height) { int start = 0; int end = height.length-1; int maxA = 0; while(start<end){ if(height[start] >= height[end]){ maxA = Math.max(maxA, (height[end]*(end-start))); int prev = height[end]; end--; while(end > start && height[end] <= prev){ end--; } } else{ maxA = Math.max(maxA, (height[start]*(end-start))); int prev = height[start]; start++; while(end > start && height[start] <= prev){ start++; } } } return maxA; } }

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