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34. Container With Most Water

mediumAsked at Workday

Given an array of heights, find two lines that hold the most water. Workday uses this for two-pointer convergence reasoning — same shape as maximizing fiscal-coverage between two timeline anchors.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Workday loops.

  • Glassdoor (2025)— Workday SDE2 phone screen.

Problem

You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]). Find two lines that, together with the x-axis, form a container that contains the most water. Return the maximum amount of water a container can store.

Constraints

  • n == height.length
  • 2 <= n <= 10^5
  • 0 <= height[i] <= 10^4

Examples

Example 1

Input
height = [1,8,6,2,5,4,8,3,7]
Output
49

Example 2

Input
height = [1,1]
Output
1

Approaches

1. All pairs brute force

Try every (i, j) and compute area.

Time
O(n^2)
Space
O(1)
let best = 0;
for (let i=0;i<height.length;i++) for (let j=i+1;j<height.length;j++) best = Math.max(best, Math.min(height[i],height[j])*(j-i));
return best;

Tradeoff: Quadratic. Fails at n=10^5.

2. Two pointers converge

lo=0, hi=n-1. Compute area; move the shorter side inward (only chance to find a taller bound).

Time
O(n)
Space
O(1)
function maxArea(height) {
  let lo = 0, hi = height.length - 1;
  let best = 0;
  while (lo < hi) {
    const h = Math.min(height[lo], height[hi]);
    best = Math.max(best, h * (hi - lo));
    if (height[lo] < height[hi]) lo++;
    else hi--;
  }
  return best;
}

Tradeoff: Single pass. Moving the shorter line is the only move that could find a taller bound — moving the taller can only narrow.

Workday-specific tips

Workday wants you to justify WHY you move the shorter pointer (not just THAT you do it). The argument: width strictly decreases, so the only way to increase area is to find a taller bound — and only the shorter side has that potential.

Common mistakes

  • Moving the taller pointer — area can only get smaller.
  • Using max instead of min for height — water level is bounded by the shorter wall.
  • Comparing heights with strict < and breaking ties wrong — both directions are fine for ties since you'll cover the case eventually.

Follow-up questions

An interviewer at Workday may pivot to one of these next:

  • Trapping Rain Water (LC 42) — same shape, different question.
  • Maximum Subarray (LC 53).
  • What if walls have width?

Solve it now

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Output

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FAQ

Why move the shorter wall?

Width is decreasing every step. If you move the taller wall, the height can only stay the same or drop — you've wasted a step. Moving the shorter wall is the only one that COULD raise the cap.

Does this work for ties?

Yes — when heights are equal, moving either is fine. You won't miss the optimal because both subproblems have the same cap and shrinking width.

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