Detect Squares: the idea to remember.

The core insight

For every stored point on the query row, use its horizontal distance as a side length and multiply the two remaining corner frequencies.

Why it works

Each nonzero horizontal partner determines exactly two possible axis aligned squares, one above and one below; frequency products count every occurrence choice.

The mistake to avoid

Repeated stored points multiply the number of squares. A zero side length is not a square.

Commented Java solution

Use java.util.* imports and a surrounding solution class. Tree snippets assume the standard TreeNode definition; linked lists use ListNode and Clone Graph uses a Node with val and neighbors; Copy Random List uses Node with val, next and random. Read the input assumptions above.

class DetectSquares {
    private final Map<Integer,Map<Integer,Integer>> rows = new HashMap<>();
    void add(int[] point) {
        rows.computeIfAbsent(point[1], ignored -> new HashMap<>()).merge(point[0],1,Integer::sum);
    }
    int count(int[] point) {
        int x = point[0], y = point[1], result = 0;
        Map<Integer,Integer> row = rows.getOrDefault(y, Collections.emptyMap());
        for (Map.Entry<Integer,Integer> other : row.entrySet()) {
            int otherX = other.getKey(), side = otherX - x;
            if (side == 0) continue;
            // Each same-row partner determines a square above and below.
            for (int otherY : new int[]{y + side, y - side}) {
                Map<Integer,Integer> parallel = rows.getOrDefault(otherY, Collections.emptyMap());
                result += other.getValue() * parallel.getOrDefault(x,0) * parallel.getOrDefault(otherX,0);
            }
        }
        return result;
    }
}

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