g0001_0100.s0011_container_with_most_water.Solution Maven / Gradle / Ivy

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package g0001_0100.s0011_container_with_most_water;

// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Array #Greedy #Two_Pointers
// #Algorithm_II_Day_4_Two_Pointers #2022_06_14_Time_2_ms_(99.81%)_Space_52.8_MB_(82.69%)

/**
 * 11 - Container With Most Water\.
 *
 * Medium
 *
 * Given `n` non-negative integers a₁, a₂, ..., a_n , where each represents a point at coordinate (i, a_i). `n` vertical lines are drawn such that the two endpoints of the line `i` is at (i, a_i) 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:**
 *
 * ![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/17/question_11.jpg)
 *
 * **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 
 *
 * **Constraints:**
 *
 * *   `n == height.length`
 * *   2 <= n <= 10⁵
 * *   0 <= height[i] <= 10⁴
**/
public class Solution {
    public int maxArea(int[] height) {
        int maxArea = -1;
        int left = 0;
        int right = height.length - 1;
        while (left < right) {
            if (height[left] < height[right]) {
                maxArea = Math.max(maxArea, height[left] * (right - left));
                left++;
            } else {
                maxArea = Math.max(maxArea, height[right] * (right - left));
                right--;
            }
        }
        return maxArea;
    }
}