Container with most water LeetCode

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.

Notice that you may not slant the container.

 Example 1:

Input: height = [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 (blue section) the container can contain is 49.

Example 2:

Input: height = [1,1]
Output: 1

Example 3:

Input: height = [4,3,2,1,4]
Output: 16

Example 4:

Input: height = [1,2,1]
Output: 2

 NOTE:

• n == height.length
• 2 <= n <= 105
• 0 <= height[i] <= 104

This problem is also popular in LeetCode and GeeksForGeeks A collection of hundreds of interview questions and solutions are available in our blog at Interview Question Solutions

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SOLUTION:

Source code: Java

	/**
	* 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.
	*/
	package com.alg.leetcode.others;
	/**
	* ref:http://rafal.io/posts/leetcode-11-container-with-most-water.html
	* @author rbaral
	*
	*/
	public class ContainerWithMaxWater {
	public int maxArea(int[] height) {
	int maxA = 0;
	int i = 0;
	int j = height.length-1;
	int curArea=0;
	while(i < j){
	if(height[i]- height[j]<=0)
	curArea = (j-i) * height[i];
	else
	curArea = (j-i) * height[j];
	maxA = Math.max(curArea,maxA);
	if(height[i] < height[j]) i++;
	else j--;
	}
	return maxA;
	}
	}

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