LeetCode - Medium - 11. Container With Most Water

Topic

• Array
• Two Pointers

Description

https://leetcode.com/problems/container-with-most-water/

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

Constraints:

• n = height.length
• 2 <= n <= 3 * 10⁴
• 0 <= height[i] <= 3 * 10⁴

Analysis

方法一：暴力演算法

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方法二：雙指標演算法

The max area is calculated by the following formula:

S = (j - i) * min(ai, aj)

We should choose (i, j) so that S is max. Note that i, j go through the range (1, n) and j > i. That’s it.

The simple way is to take all possibilities of (i, j) and compare all obtained S. The time complexity is n * (n-1) / 2.（暴力演算法）

What we gonna do is to choose all possibilities of (i, j) in a wise way. I noticed that many submitted solutions here can’t explain why when :

• ai < aj we will check the next (i+1, j) (or move i to the right)
• ai >= aj we will check the next (i, j-1) (or move j to the left)

（也就是保留兩者中最大的下標，移動最小值的下標（i的向右移動，j的向左移動））

Here is the explaination for that:

• When ai < aj , we don’t need to calculate all (i, j-1), (i, j-2), … Why? because these max areas are smaller than our S at (i, j)

Assume at (i, j-1) we have

ai < aj
S = (j - i) * min(ai, aj) = (j - i) * ai

if aj-1 >= ai
S' = (j - 1 - i) * min(ai, aj-1) = (j - 1 - i) * ai
S > S'

if aj-1 < ai
S' = (j - 1 - i) * min(ai, aj-1) = (j - 1 - i) * aj-1
S > S'

So no matter aj-1 >= ai or aj-1 < ai，S' < S always true

we don’t need to calculate Similar at (i, j-2), (i, j-3), etc.

So, that’s why when ai < aj, we should check the next at (i+1, j) (or move i to the right)

• When ai >= aj, the same thing, all (i+1, j), (i+2, j), … are not needed to calculate.

ai >= aj
S = (j - i) * min(ai, aj) = (j - i) * aj

if ai+1 >= aj
S' = (j - (i + 1)) * min(ai+1, aj)
< (j - i - 1) * aj
S' < S

if ai+1 < aj
S' = (j - (i + 1)) * min(ai+1, aj)
< (j - i - 1) * ai+1
S' < S

So no matter ai+1 >= aj or ai+1 < aj，S' < S always true

We should check the next at (i, j-1) (or move j to the left)

PS.似非而是的解法啊！

Submission

public class ContainerWithMostWater {

	// 方法一：暴力演算法
	public int maxArea1(int[] height) {
		int maxCapacity = 0;

		for (int i = 0; i < height.length; i++) {
			for (int j = i + 1; j < height.length; j++) {
				int min = Math.min(height[i], height[j]);
				int distance = j - i;
				maxCapacity = Math.max(maxCapacity, min * distance);
			}
		}
		return maxCapacity;
	}

	// 方法二：雙指標
	public int maxArea2(int[] height) {
		int S = 0, i = 0, j = height.length - 1;
		while (i < j) {
			S = Math.max(S, (j - i) * Math.min(height[i], height[j]));
			if (height[i] < height[j])
				i++;
			else
				j--;
		}
		return S;
	}

}

Test

import static org.junit.Assert.*;
import org.junit.Test;

public class ContainerWithMostWaterTest {

	@Test
	public void test() {
		ContainerWithMostWater obj = new ContainerWithMostWater();

		assertEquals(49, obj.maxArea1(new int[] {1, 8, 6, 2, 5, 4, 8, 3, 7}));
		assertEquals(1, obj.maxArea1(new int[] {1, 1}));
		assertEquals(16, obj.maxArea1(new int[] {4, 3, 2, 1, 4}));
		assertEquals(2, obj.maxArea1(new int[] {1, 2, 1}));
		
		assertEquals(49, obj.maxArea2(new int[] {1, 8, 6, 2, 5, 4, 8, 3, 7}));
		assertEquals(1, obj.maxArea2(new int[] {1, 1}));
		assertEquals(16, obj.maxArea2(new int[] {4, 3, 2, 1, 4}));
		assertEquals(2, obj.maxArea2(new int[] {1, 2, 1}));
	}
}