#11 Container With Most Water

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the problem

In plain words

height[i] is a vertical wall at x = i. Pick two walls that, with the x-axis, hold the most water. Return that maximum area.

in height = [1,8,6,2,5,4,8,3,7]

out 49

Walls at index 1 (8) and 8 (7): min(8,7) × 7 = 49.

in height = [1,1]

out 1

2 ≤ n ≤ 10⁵
0 ≤ height[i] ≤ 10⁴

your turn

Try it before you peek

Read the examples, say the brute force out loud, then write code: here or on LeetCode ↗. Stuck for 20 minutes? Open one hint at a time.

Solve it in the browser

LeetCode-style: complete the class. Compiled by the real javac (Java 17) and judged right here — nothing is sent anywhere.

Solutions are hidden so you can try first. Even 20 minutes of real effort makes the answer stick far better than reading it.

Marking the problem solved, or opening every hint, also shows it.

the aha

Key insight

patternTwo pointers (opposite ends)

Area = width × shorter wall. Start with maximum width. Moving the taller wall can't raise the shorter wall and only shrinks width — so always move the shorter one.

solutions

From brute force to optimal

Compute the area for all pairs.

Two loops, area = (j − i) × min(h[i], h[j]).

timeO(n²)

All pairs.

spaceO(1)

One running max.

java

1class Solution {2    public int maxArea(int[] height) {3        int best = 0;4        for (int i = 0; i < height.length; i++)5            for (int j = i + 1; j < height.length; j++)6                best = Math.max(best, (j - i) * Math.min(height[i], height[j]));7        return best;8    }9}

Start widest; after measuring, discard the shorter wall.

left = 0, right = n−1.
best = max(best, width × min height).
Move the pointer at the shorter wall inward.

timeO(n)

Each step discards one wall.

spaceO(1)

Two pointers and a max.

java

1class Solution {2    public int maxArea(int[] height) {3        int left = 0, right = height.length - 1, best = 0;4        while (left < right) {5            int h = Math.min(height[left], height[right]);6            best = Math.max(best, h * (right - left));7            // The shorter wall limits the water. Moving the taller wall can only8            // shrink the width without raising the limit — so move the shorter one.9            if (height[left] < height[right]) left++;10            else right--;11        }12        return best;13    }14}

sim

Dry run

loading simulation…

java

1class Solution {2    public int maxArea(int[] height) {3        int left = 0, right = height.length - 1, best = 0;4        while (left < right) {5            int h = Math.min(height[left], height[right]);6            best = Math.max(best, h * (right - left));7            // The shorter wall limits the water. Moving the taller wall can only8            // shrink the width without raising the limit — so move the shorter one.9            if (height[left] < height[right]) left++;10            else right--;11        }12        return best;13    }14}

in the room

How to say it in the interview

talk track

Prove the move: "Any container that keeps the shorter wall is narrower and no taller, so it can't beat what we just measured."

traps

Confusing this with Trapping Rain Water — here only two walls matter; walls in between don't block anything.

what they ask next

Follow-ups

Why is it safe to move either pointer when both walls are equal?

Any pair that keeps one of them is narrower and still capped at that same height, so neither wall can do better. Dropping either loses nothing.

Return the two indices instead of the area?

Same scan — whenever you update best, also save left and right. Still O(n) time, O(1) space.

before you submit

Edge cases to test

Two walls only: the answer is min(h[0], h[1]) × 1.
Zero-height walls hold nothing: [0, 0] → 0.
Equal walls at both ends, like [4, 3, 2, 1, 4] → 16: moving either side is safe.
10⁵ walls: every pair is about 5·10⁹ areas; two pointers is at most 10⁵ − 1 steps.

practise next