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LeetCode Solutions

378. Kth Smallest Element in a Sorted Matrix

Time: $O(x + k\log x)$, where $x = \min(n, k)$

Space: $O(x)$, where $x = \min(n, k)$

			


struct T {
  int i;
  int j;
  int num;  // matrix[i][j]
  T(int i, int j, int num) : i(i), j(j), num(num) {}
};

class Solution {
 public:
  int kthSmallest(vector<vector<int>>& matrix, int k) {
    auto compare = [&](const T& a, const T& b) { return a.num > b.num; };
    priority_queue<T, vector<T>, decltype(compare)> minHeap(compare);

    for (int i = 0; i < k && i < matrix.size(); ++i)
      minHeap.emplace(i, 0, matrix[i][0]);

    while (k-- > 1) {
      const auto [i, j, _] = minHeap.top();
      minHeap.pop();
      if (j + 1 < matrix[0].size())
        minHeap.emplace(i, j + 1, matrix[i][j + 1]);
    }

    return minHeap.top().num;
  }
};




			


class T {
  public int i;
  public int j;
  public int num; // matrix[i][j]
  public T(int i, int j, int num) {
    this.i = i;
    this.j = j;
    this.num = num;
  }
}

class Solution {
  public int kthSmallest(int[][] matrix, int k) {
    Queue<T> minHeap = new PriorityQueue<>((a, b) -> a.num - b.num);

    for (int i = 0; i < k && i < matrix.length; ++i)
      minHeap.offer(new T(i, 0, matrix[i][0]));

    while (k-- > 1) {
      final int i = minHeap.peek().i;
      final int j = minHeap.poll().j;
      if (j + 1 < matrix[0].length)
        minHeap.offer(new T(i, j + 1, matrix[i][j + 1]));
    }

    return minHeap.peek().num;
  }
}




			


class Solution:
  def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
    minHeap = []  # (matrix[i][j], i, j)

    i = 0
    while i < k and i < len(matrix):
      heapq.heappush(minHeap, (matrix[i][0], i, 0))
      i += 1

    while k > 1:
      k -= 1
      _, i, j = heapq.heappop(minHeap)
      if j + 1 < len(matrix[0]):
        heapq.heappush(minHeap, (matrix[i][j + 1], i, j + 1))

    return minHeap[0][0]