Count Islands After Land Additions
Reported by candidates from Google's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
This Google OA, reported in September 2026, looks like a grid problem but it's really a connectivity-counting problem in disguise. Land gets added one cell at a time, and after each add you report the island count. It's Union-Find. If you try to rerun a flood fill after every addition, you'll die on 100000 positions over a grid up to 1000 by 1000. If the pattern slips your mind mid-assessment, StealthCoder runs invisibly as a safety net and gives you the solution live.
The problem
Start with a rows by cols grid containing only water. Process each position in positions in order by turning that cell into land. After every addition, return the number of islands. An island is a maximal group of land cells connected vertically or horizontally. Adding a cell that is already land changes nothing but still produces an output. Function countIslandsAfterAdditions(rows: int, cols: int, positions: int[][]) → int[] Examples Example 1 rows = 3 cols = 3 positions = [[0,0],[0,1],[1,2],[2,1],[1,1]] return = [1,1,2,3,1] The final center cell joins the three existing components into one island. Example 2 rows = 1 cols = 3 positions = [[0,1],[0,1],[0,0],[0,2]] return = [1,1,1,1] The repeated addition is a no-op, and both later cells attach to the existing island. Example 3 rows = 2 cols = 2 positions = [[0,0],[1,1],[0,1],[1,0]] return = [1,2,1,1] Diagonal cells are separate. Adding either remaining edge-adjacent cell merges them. Constraints 1 <= rows, cols <= 1000 1 <= positions.length <= 100000 Every position is [row, col] with 0 <= row < rows and 0 <= col < cols. Positions may repeat.
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick: keep a count and a union-find over cells indexed as r*cols+c. When you add a new land cell, increment count by one. Then check its four neighbors. Every neighbor that's already land and sits in a different set triggers a union and decrements count by one. Output count after each step. The pitfall is duplicates. If a position is already land, skip all work and just append the current count, or you'll double count. Another miss is allocating a full 1000x1000 DSU eagerly and re-scanning the grid. Use path compression and union by size or rank so each op is near constant. Total cost is about O(k * alpha(n)) for k positions. Diagonals don't connect, so only check up, down, left, right. StealthCoder is your hedge if the DSU boilerplate or the duplicate edge case escapes you while the clock runs.
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Count Islands After Land Additions FAQ
What's the trick to the Google island additions problem?+
Use Union-Find, not repeated DFS. Each new land cell adds one island, then every distinct adjacent land component it merges with subtracts one. Track the running count and append it after each position. That's the whole idea.
How do I handle repeated positions?+
Keep a land marker, like a boolean array or set. If the cell is already land, don't touch the count or unions. Just push the current count to the result. Example 2 tests exactly this, where the repeated [0,1] keeps the answer at 1.
Why not run DFS or BFS after every addition?+
With up to 100000 positions on a grid up to 1000 by 1000, a full scan each time is far too slow. Union-Find updates only the four neighbors of the new cell, so each step is nearly constant time.
Do diagonal cells count as connected?+
No. Only vertical and horizontal neighbors connect. Example 3 shows it: [0,0] and [1,1] stay as two islands until an edge-adjacent cell joins them. Check just four directions when unioning.
How do I prepare for this in 48 hours?+
Write a clean Union-Find with path compression and union by size from memory. Then code this problem once, using index r*cols+c and a land array. Test the three given examples, especially the duplicate and the final merge of three components.