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1762. Buildings With an Ocean View

mediumAsked at Meta

Given building heights, return indices that can see the ocean to the right. Meta asks this to test whether you reach for the right-to-left sweep that's O(n), not the O(n^2) per-building lookahead.

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

Source citations

Public interview reports confirming this problem appears in Meta loops.

  • Glassdoor (2026-Q1)— Meta E3/E4 onsite reports cite this as a Meta-favorite warmup.
  • Blind (2025-12)— Recurring Meta phone-screen problem.

Problem

There are n buildings in a line. You are given an integer array heights of size n that represents the heights of the buildings in the line. The ocean is to the right of the buildings. A building has an ocean view if the building can see the ocean without obstructions. Formally, a building has an ocean view if all the buildings to its right have a smaller height. Return a list of indices (0-indexed) of buildings that have an ocean view, sorted in increasing order.

Constraints

  • 1 <= heights.length <= 10^5
  • 1 <= heights[i] <= 10^9

Examples

Example 1

Input
heights = [4,2,3,1]
Output
[0,2,3]

Example 2

Input
heights = [4,3,2,1]
Output
[0,1,2,3]

Example 3

Input
heights = [1,3,2,4]
Output
[3]

Approaches

1. Brute-force lookahead per index

For each i, check whether all heights[j] for j > i are strictly less than heights[i].

Time
O(n^2)
Space
O(1)
function findBuildingsBrute(heights) {
  const result = [];
  for (let i = 0; i < heights.length; i++) {
    let canSee = true;
    for (let j = i + 1; j < heights.length; j++) {
      if (heights[j] >= heights[i]) { canSee = false; break; }
    }
    if (canSee) result.push(i);
  }
  return result;
}

Tradeoff: Quadratic — fails at the upper bound. Mention only to anchor the optimal.

2. Right-to-left sweep with running max (optimal)

Walk from right to left. Track the max height seen so far. A building has ocean view iff its height > current max.

Time
O(n)
Space
O(n) for output
function findBuildings(heights) {
  const result = [];
  let maxRight = 0;
  for (let i = heights.length - 1; i >= 0; i--) {
    if (heights[i] > maxRight) {
      result.push(i);
      maxRight = heights[i];
    }
  }
  return result.reverse();
}

Tradeoff: Linear time. The right-to-left direction is the key — every building only needs to compare against the max of everything to its right, which we accumulate in one variable.

Meta-specific tips

Meta interviewers grade this on whether you recognize the right-to-left direction. State out loud: 'I'm walking right-to-left because every building's ocean view depends only on what's to its right.' This problem is a Meta favorite specifically because the wrong-direction sweep doesn't work — testing whether you can identify the natural processing direction.

Common mistakes

  • Sweeping left-to-right, then needing extra storage for the suffix max.
  • Forgetting to reverse the result at the end (we pushed indices in reverse order).
  • Using >= instead of > — the problem says strictly smaller, so equal heights block the view.

Follow-up questions

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

  • What if the ocean were on both sides?
  • What if buildings could be in 2D (towers in a city)?
  • What if heights could change online (insertions)?

Solve it now

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Output

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FAQ

Why right-to-left and not left-to-right?

Each building's ocean view depends only on the max of the buildings TO ITS RIGHT. Right-to-left lets us accumulate that max in O(1) extra space.

Could we do it left-to-right?

Yes — precompute a suffix max array (O(n) space), then sweep left-to-right. Same time complexity but more space. Right-to-left wins on elegance.

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