Container With Most Water

mediumTwo Pointers44 stepsJava · Python · C++

Pick two lines that hold the most water: area = min(height) × width. Start with the widest pair and always move the shorter line inward, since keeping it can never produce a bigger area.

▶ Watch the animated trace

Traced input: height = [1, 8, 6, 2, 5, 4, 8, 3, 7] → 49. Original problem: neetcode.io/problems/max-water-container

Complexity

BestO(n)

AverageO(n)

WorstO(n)

SpaceO(1)

Each step discards one line.

Key idea

The shorter line limits the area. Any container that keeps it and is narrower is strictly worse.
So moving the shorter pointer is the only move that might find a bigger area.
Track the best area seen along the way.

Common mistakes

Moving the taller line can only lose area.
Checking every pair is O(n²) and times out.
Width is r - l, not r - l + 1.

Solution

Java

int maxArea(int[] height) {
    int l = 0, r = height.length - 1, best = 0;
    while (l < r) {
        int area = Math.min(height[l], height[r]) * (r - l);
        best = Math.max(best, area);
        if (height[l] < height[r]) l++;
        else r--;
    }
    return best;
}

Python

def maxArea(height: List[int]) -> int:
    l, r, best = 0, len(height) - 1, 0
    while l < r:
        area = min(height[l], height[r]) * (r - l)
        best = max(best, area)
        if height[l] < height[r]: l += 1
        else: r -= 1
    return best

C++

int maxArea(vector<int>& height) {
    int l = 0, r = (int)height.size() - 1, best = 0;
    while (l < r) {
        int area = min(height[l], height[r]) * (r - l);
        best = max(best, area);
        if (height[l] < height[r]) l++;
        else r--;
    }
    return best;
}

Edge cases

Each one has its own trace in the interactive view.

Case	Input	Answer
Two lines One pair, width 1.	`height = [1, 1]`	`1`
Equal heights The widest pair wins.	`height = [4, 4, 4, 4]`	`12`
Increasing heights Best is 2 × 3 or 3 × 2.	`height = [1, 2, 3, 4, 5]`	`6`
Zero heights Every pair includes a line of height 0, so nothing is held.	`height = [0, 2, 0]`	`0`