Solution: Container With Most Water

Leetcode Solutions (161 Part Series)

1 Solution: Next Permutation
2 Solution: Trim a Binary Search Tree
… 157 more parts…
3 Leetcode Solutions Index
4 Solution: Minimize Deviation in Array
5 Solution: Vertical Order Traversal of a Binary Tree
6 Solution: Count Ways to Make Array With Product
7 Solution: Smallest String With A Given Numeric Value
8 Solution: Linked List Cycle
9 Solution: Path With Minimum Effort
10 Solution: Concatenation of Consecutive Binary Numbers
11 Solution: Minimum Operations to Make a Subsequence
12 Solution: Longest Harmonious Subsequence
13 Solution: Simplify Path
14 Solution: Building Boxes
15 Solution: Decode XORed Permutation
16 Solution: Binary Tree Right Side View
17 Solution: Find Kth Largest XOR Coordinate Value
18 Solution: Change Minimum Characters to Satisfy One of Three Conditions
19 Solution: Shortest Distance to a Character
20 Solution: Peeking Iterator
21 Solution: Convert BST to Greater Tree
22 Solution: Copy List with Random Pointer
23 Solution: Valid Anagram
24 Solution: Number of Steps to Reduce a Number to Zero
25 Solution: Shortest Path in Binary Matrix
26 Solution: Is Graph Bipartite?
27 Solution: Maximum Score From Removing Substrings (ver. 1)
28 Solution: Maximum Score From Removing Substrings (ver. 2)
29 Solution: Sort the Matrix Diagonally
30 Solution: The K Weakest Rows in a Matrix (ver. 1)
31 Solution: The K Weakest Rows in a Matrix (ver. 2)
32 Solution: Letter Case Permutation
33 Solution: Container With Most Water
34 Solution: Arithmetic Slices
35 Solution: Minimum Remove to Make Valid Parentheses
36 Solution: Roman to Integer
37 Solution: Broken Calculator
38 Solution: Find the Most Competitive Subsequence
39 Solution: Longest Word in Dictionary through Deleting
40 Solution: Search a 2D Matrix II
41 Solution: Score of Parentheses
42 Solution: Shortest Unsorted Continuous Subarray
43 Solution: Validate Stack Sequences
44 Solution: Divide Two Integers (ver. 1)
45 Solution: Divide Two Integers (ver. 2)
46 Solution: Maximum Frequency Stack
47 Solution: Distribute Candies
48 Solution: Set Mismatch (ver. 1)
49 Solution: Set Mismatch (ver. 2)
50 Solution: Missing Number
51 Solution: Intersection of Two Linked Lists
52 Solution: Average of Levels in Binary Tree
53 Solution: Short Encoding of Words (ver. 1)
54 Solution: Design HashMap (ver. 1)
55 Solution: Short Encoding of Words (ver. 2)
56 Solution: Design HashMap (ver. 2)
57 Solution: Remove Palindromic Subsequences
58 Solution: Add One Row to Tree
59 Solution: Integer to Roman
60 Solution: Coin Change
61 Solution: Check If a String Contains All Binary Codes of Size K
62 Solution: Binary Trees With Factors
63 Solution: Swapping Nodes in a Linked List
64 Solution: Encode and Decode TinyURL
65 Solution: Best Time to Buy and Sell Stock with Transaction Fee
66 Solution: Generate Random Point in a Circle
67 Solution: Wiggle Subsequence
68 Solution: Keys and Rooms
69 Solution: Design Underground System
70 Solution: Reordered Power of 2
71 Solution: Vowel Spellchecker
72 Solution: 3Sum With Multiplicity
73 Solution: Advantage Shuffle
74 Solution: Pacific Atlantic Water Flow
75 Solution: Word Subsets
76 Solution: Palindromic Substrings
77 Solution: Reconstruct Original Digits from English
78 Solution: Flip Binary Tree To Match Preorder Traversal
79 Solution: Russian Doll Envelopes
80 Solution: Stamping The Sequence
81 Solution: Palindrome Linked List
82 Solution: Ones and Zeroes
83 Solution: Longest Valid Parentheses
84 Solution: Design Circular Queue
85 Solution: Global and Local Inversions
86 Solution: Minimum Operations to Make Array Equal
87 Solution: Determine if String Halves Are Alike
88 Solution: Letter Combinations of a Phone Number
89 Solution: Verifying an Alien Dictionary
90 Solution: Longest Increasing Path in a Matrix
91 Solution: Deepest Leaves Sum
92 Solution: Beautiful Arrangement II
93 Solution: Flatten Nested List Iterator
94 Solution: Partition List
95 Solution: Fibonacci Number
96 Solution: Remove All Adjacent Duplicates in String II
97 Solution: Number of Submatrices That Sum to Target
98 Solution: Remove Nth Node From End of List
99 Solution: Combination Sum IV
100 Solution: N-ary Tree Preorder Traversal
101 Solution: Triangle
102 Solution: Brick Wall
103 Solution: Count Binary Substrings
104 Solution: Critical Connections in a Network
105 Solution: Rotate Image
106 Solution: Furthest Building You Can Reach
107 Solution: Power of Three
108 Solution: Unique Paths II
109 Solution: Find First and Last Position of Element in Sorted Array
110 Solution: Powerful Integers
111 Solution: Prefix and Suffix Search
112 Solution: Course Schedule III
113 Solution: Running Sum of 1d Array
114 Solution: Non-decreasing Array
115 Solution: Jump Game II
116 Solution: Convert Sorted List to Binary Search Tree
117 Solution: Delete Operation for Two Strings
118 Solution: Super Palindromes
119 Solution: Construct Target Array With Multiple Sums
120 Solution: Count Primes
121 Solution: Maximum Points You Can Obtain from Cards
122 Solution: Range Sum Query 2D – Immutable
123 Solution: Ambiguous Coordinates
124 Solution: Flatten Binary Tree to Linked List
125 Solution: Valid Number
126 Solution: Binary Tree Cameras
127 Solution: Longest String Chain
128 Solution: Find Duplicate File in System
129 Solution: Minimum Moves to Equal Array Elements II
130 Solution: Binary Tree Level Order Traversal
131 Solution: Find and Replace Pattern
132 Solution: N-Queens
133 Solution: To Lower Case
134 Solution: Evaluate Reverse Polish Notation
135 Solution: Partitioning Into Minimum Number Of Deci-Binary Numbers
136 Solution: Maximum Product of Word Lengths
137 Solution: Maximum Erasure Value
138 Solution: N-Queens II
139 Solution: Maximum Gap
140 Solution: Search Suggestions System
141 Solution: Max Area of Island
142 Solution: Interleaving String
143 Solution: Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts
144 Solution: Open the Lock
145 Solution: Maximum Performance of a Team
146 Solution: Longest Consecutive Sequence
147 Solution: Min Cost Climbing Stairs
148 Solution: Construct Binary Tree from Preorder and Inorder Traversal
149 Solution: Jump Game VI
150 Solution: My Calendar I
151 Solution: Stone Game VII
152 Solution: Minimum Number of Refueling Stops
153 Solution: Palindrome Pairs
154 Solution: Maximum Units on a Truck
155 Solution: Matchsticks to Square
156 Solution: Generate Parentheses
157 Solution: Number of Subarrays with Bounded Maximum
158 Solution: Swim in Rising Water
159 Solution: Pascal’s Triangle
160 Solution: Out of Boundary Paths
161 Solution: Redundant Connection

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode’s forums.

Leetcode Problem #11 (Medium): Container With Most Water

Description:

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.

Examples:

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.
Visual:

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 * 10^4
0 <= height[i] <= 3 * 10^4

Idea:

The first thing we should realize is that the amount of water contained is always going to be a rectangle whose area is defined as length * width. The width of any container will be the difference between the index of the two lines (i and j), and the height will be whichever of the two sides is the lowest (min(H[i], H[j])).

The brute force approach would be to compare every single pair of indexes in H, but that would be far too slow. Instead, we can observe that if we start with the lines on the opposite ends and move inward, the only possible time the area could be larger is when the height increases, since the width will continuously get smaller.

This is very easily observed with the use of visuals. Let’s say we start with a graph of H like this:

The first step would be to find our starting container described by the lines on either end:

We can tell that the line on the right end will never make a better match, because any further match would have a smaller width and the container is already the maximum height that that line can support. That means that our next move should be to slide j to the left and pick a new line:

This is a clear improvement over the last container. We only moved over one line, but we more than doubled the height. Now, it’s the line on the left end that’s the limiting factor, so the next step will be to slide i to the right. Just looking at the visual, however, it’s obvious that we can skip the next few lines because they’re already underwater, so we should go to the first line that’s larger than the current water height:

This time, it doesn’t look like we made much of a gain, despite the fact that the water level rose a bit, because we lost more in width than we made up for in height. That means that we always have to check at each new possible stop to see if the new container area is better than the current best. Just lik before we can slide j to the left again:

This move also doesn’t appear to have led to a better container. But here we can see that it’s definitely possible to have to move the same side twice in a row, as the j line is still the lower of the two:

This is obviously the last possible container to check, and like the last few before it, it doesn’t appear to be the best match. Still, we can understand that it’s entirely possible for the best container in a different example to be only one index apart, if both lines are extremely tall.

Putting together everything, it’s clear that we need to make a 2-pointer sliding window solution. We’ll start from either end and at each step we’ll check the container area, then we’ll shift the lower-valued pointer inward. Once the two pointers meet, we know that we must have exhausted all possible containers and we should return our answer (ans).

Implementation:

Javascript was weirdly more performant when using both Math.max() and Math.min() rather than performing more basic comparisons, even with duplicated effort in the ternary.

For the other languages, it made more sense (and was ultimately more performant) to only have to do the basic comparisons once each.

Javascript Code:



var maxArea = function(H) {
    let ans = 0, i = 0, j = H.length-1
    while (i < j) {
        ans = Math.max(ans, Math.min(H[i], H[j]) * (j - i))
        H[i] <= H[j] ? i++ : j--
    }
    return ans
};

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Python Code:



class Solution:
    def maxArea(self, H: List[int]) -> int:
        ans, i, j = 0, 0, len(H)-1
        while (i < j):
            if H[i] <= H[j]:
                res = H[i] * (j - i)
                i += 1
            else:
                res = H[j] * (j - i)
                j -= 1
            if res > ans: ans = res
        return ans

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Java Code:



class Solution {
    public int maxArea(int[] H) {
        int ans = 0, i = 0, j = H.length-1, res = 0;
        while (i < j) {
            if (H[i] <= H[j]) {
                res = H[i] * (j - i);
                i++;
            } else {
                res = H[j] * (j - i);
                j--;
            }
            if (res > ans) ans = res;
        }
        return ans;
    }
}

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C++ Code:



class Solution {
public:
    int maxArea(vector<int>& H) {
        int ans = 0, i = 0, j = H.size()-1, res = 0;
        while (i < j) {
            if (H[i] <= H[j]) {
                res = H[i] * (j - i);
                i++;
            } else {
                res = H[j] * (j - i);
                j--;
            }
            if (res > ans) ans = res;
        }
        return ans;
    }
};

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Leetcode Solutions (161 Part Series)

原文链接：Solution: Container With Most Water

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