The O(n) Club: Most Stones Removed with Same Row or Column: Stone-Crushing Group Therapy

The O(n) Club: Most Stones Removed with Same Row or Column — Stone-Crushing Group Therapy

⚡ TL;DR

Imagine every stone is a bored dev at a hackathon: anyone sharing a row or column can leave as a group, but one poor soul from each clique is stuck ordering pizza. Brute force? It’s the fast track to a time-limit-exceeded meltdown:

// Painful, slow, and not recommended
int res = 0;
for (int i = 0; i < stones.length; i++) {
    for (int j = 0; j < stones.length; j++) {
        if (i != j && (stones[i][0] == stones[j][0] || stones[i][1] == stones[j][1])) {
            // Try to remove stones[i] here
        }
    }
}
// Spoiler: This will have you regretting choosing computer science.

🧠 How Devs Usually Mess This Up

Everyone’s first instinct? Remove stones in pairs over and over, maybe throw in a 1000-line recursion. But stones don’t just connect to neighbors—they join distant cousins through shared rows and columns. Also, you can’t remove the final stone in a group, unless you’re a villain in a disaster movie. Please don’t be that villain.

🔍 What’s Actually Going On

This is less “pair removal” and more “let’s find the secret party rooms.” Stones in the same row or column form one raucous connected component. Within each group, you can kick out every stone but one (so that last stone has to finish cleaning up all the chips). The trick is to count the number of groups (connected components). Answer: stones.length - groups. Union-Find (DSU) is the bouncer at this party, efficiently telling you who’s in which clique. If you see anyone trying brute force, feel free to send them this link (and coffee).

🛠️ PseudoCode

• Model each stone as a node. Two stones are friends if they share a row or column.
• Union-find setup: For each stone, union(stone[0], stone[1] + 10000) (offset to avoid row/column mixups).
• Track unique groups using a set to collect parent nodes.
• Final answer: result = stones.length - groupCount

• Step-by-step:
  • For each stone, join (union) its row and column (with offset).
  • After processing all stones, for each one, store its root (find) in a set (for unique group count).
  • Number of stones you can remove: total stones minus the number of unique groups.

💻 The Code

// Union-Find for the stone party
class Solution {
    Map<integer integer> parent = new HashMap<>();
     public int removeStones(int[][] stones) {
        for (int[] stone : stones) {
            // Offset column value to separate from row
            union(stone[0], stone[1] + 10000);
        }
        Set
<integer> groups = new HashSet<>();
        for (int[] stone : stones) {
            groups.add(find(stone[0]));
        }
        return stones.length - groups.size();
    }
     private int find(int x) {
        parent.putIfAbsent(x, x);
        if (parent.get(x) != x) {
            parent.put(x, find(parent.get(x)));
        }
        return parent.get(x);
    }
     private void union(int x, int y) {
        parent.putIfAbsent(x, x);
        parent.putIfAbsent(y, y);
        parent.put(find(x), find(y));
    }
}
</integer></integer>

⚠️ Pitfalls, Traps & Runtime Smacks

• Isolated Stones: Stones not sharing a row or column with anyone must stay (they’re the lonely ones—respect their privacy).
• Indirect Chains: If A shares row with B, and B shares column with C, then A and C are BFFs—don’t forget that chain of trust.
• Time Costs: Union-Find is almost linear (like a caffeine-fueled dev sprint); DFS/BFS can get slow real fast if you try to visit every possible connection.
• Row/Col Key Collisions: Always offset columns by a huge number—unless you like bugs where row 1 = col 1 (no one likes those bugs).

🧠 Interviewer Brain-Tattoo

“Stop thinking about removing pairs. Start thinking like a nightclub bouncer with a Union-Find spreadsheet—groups rule, singles drool.”