Sort Every Matrix Border Layer
Reported by candidates from Capital One's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
Capital One reported this one in July 2026, and the 200 by 200 limit is the first thing to read. That's 40,000 cells, so nothing fancy is needed, but a sloppy approach that re-scans the grid per layer or double-visits corners will bite you. The task is to sort each concentric border ring on its own and write it back clockwise. The hinted pattern says BFS, but this is really matrix traversal and sorting. If you blank on the corner rules mid-assessment, StealthCoder is the quiet hedge sitting on your desktop.
The problem
Given a rectangular integer matrix, sort the values on every concentric border layer independently in ascending order. For each layer, visit its cells clockwise beginning at that layer's top-left cell: Traverse the top edge from left to right. Traverse the right edge from top to bottom, excluding the already visited top-right cell. If the layer has more than one row, traverse the bottom edge from right to left, excluding the already visited bottom-right cell. If the layer has more than one column, traverse the left edge from bottom to top, excluding both already visited corner cells. Collect the values in this order, sort them in ascending order, and write them back along the same clockwise path. Continue from the outermost layer toward the center. A layer consisting of one row, one column, or one cell follows the same rule without visiting any cell twice. Return the matrix after every layer has been sorted. Function sortMatrixBorderLayers(matrix: int[][]) → int[][] Examples Example 1 matrix = [[4,1,3],[2,9,8],[7,6,5]] return = [[1,2,3],[8,9,4],[7,6,5]] The outer layer is read as [4,1,3,8,5,6,7,2]. Writing its sorted values [1,2,3,4,5,6,7,8] along the same path produces the returned matrix. The center value 9 forms a one-cell layer and remains unchanged. Example 2 matrix = [[8,1,6,2],[7,4,5,3]] return = [[1,2,3,4],[8,7,6,5]] This two-row matrix has one layer. Its clockwise order is [8,1,6,2,3,5,4,7], so the values 1 through 8 are written in ascending order around that path. Example 3 matrix = [[3],[1],[2]] return = [[1],[2],[3]] A single-column layer is visited from top to bottom exactly once, then sorted in that order. Constraints 1 <= matrix.length <= 200 1 <= matrix[i].length <= 200 Every row has the same length. -10^9 <= matrix[i][j] <= 10^9
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick is to build the clockwise coordinate list for each layer once, then reuse it for both reading and writing. Pull values by those coordinates, sort, and assign back in the same order. Total work is about O(n*m log(n+m)), well within 40,000 cells. The real pitfall is the degenerate layers. When a layer has one row or one column, you must skip the bottom and left edges, or you'll visit cells twice and corrupt the sort. Loop layers while top <= bottom and left <= right. Check Example 3, the single column, and Example 2, the two-row case, before you submit. Don't bother with BFS, there's no graph here. If the corner logic slips under time pressure, StealthCoder can hand you a clean traversal in the live OA without anyone seeing it.
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Sort Every Matrix Border Layer FAQ
What's the trick in Sort Every Matrix Border Layer?+
Generate the clockwise coordinates for each layer, read values at those positions, sort them, and write them back in the same order. Separating path generation from sorting keeps the code simple and makes the one-row and one-column edge cases the only thing you have to get right.
Is this really a BFS problem?+
No. The hinted pattern says BFS, but there's no graph or shortest path. It's matrix layer traversal plus sorting. Thinking in rings, not neighbors, is the fastest route to a working solution.
What edge cases break most solutions?+
Layers with a single row, a single column, or a single cell. If you always run all four edge loops, you'll double-count cells. Guard the bottom edge with more than one row and the left edge with more than one column, exactly as the statement says.
How fast does it need to be?+
With up to 200 by 200 cells, collecting and sorting each layer is plenty fast. Total cost is dominated by sorting ring values, roughly O(nm log(n+m)). You don't need anything cleverer than that.
How do I prepare for this in 48 hours?+
Practice spiral-order traversal until the four-edge loop with boundary checks is automatic. Then hand-trace the three examples, especially the single column [[3],[1],[2]]. That covers nearly every way this problem can go wrong.