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[8/2] 378. Kth Smallest Element in a Sorted Matrix

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  • 학부유학생 작성
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본문

Given an n x n matrix where each of the rows and columns is sorted in ascending order, return the kth smallest element in the matrix.

Note that it is the kth smallest element in the sorted order, not the kth distinct element.

You must find a solution with a memory complexity better than O(n2).

 

Example 1:

Input: matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8
Output: 13
Explanation: The elements in the matrix are [1,5,9,10,11,12,13,13,15], and the 8th smallest number is 13

Example 2:

Input: matrix = [[-5]], k = 1
Output: -5

 

Constraints:

  • n == matrix.length == matrix[i].length
  • 1 <= n <= 300
  • -109 <= matrix[i][j] <= 109
  • All the rows and columns of matrix are guaranteed to be sorted in non-decreasing order.
  • 1 <= k <= n2

 

Follow up:

  • Could you solve the problem with a constant memory (i.e., O(1) memory complexity)?
  • Could you solve the problem in O(n) time complexity? The solution may be too advanced for an interview but you may find reading this paper fun.
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학부유학생님의 댓글

  • 학부유학생
  • 작성일
Runtime: 314 ms, faster than 47.76% of Python3 online submissions for Kth Smallest Element in a Sorted Matrix.
Memory Usage: 18.7 MB, less than 38.85% of Python3 online submissions for Kth Smallest Element in a Sorted Matrix.

import heapq
class Solution:
    def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
        min_heap = [(matrix[i][0], i, 0) for i in range(len(matrix)) ]
        
        heapq.heapify(min_heap)
        
        res = min_heap[0][0]
        for _ in range(k):
            res, i, j = heapq.heappop(min_heap)
            if j + 1< len(matrix[0]):
                heapq.heappush(min_heap,(matrix[i][j+1],i,j+1))
        
        return res

mingki님의 댓글

  • mingki
  • 작성일
C++
Runtime: 239 ms, faster than 5.03% of C++ online submissions for Kth Smallest Element in a Sorted Matrix.
Memory Usage: 17 MB, less than 7.97% of C++ online submissions for Kth Smallest Element in a Sorted Matrix.
class Solution {
public:
    int kthSmallest(vector<vector<int>>& matrix, int k) {
        int n = matrix.size();
        vector<int> indice(n, 0);
        priority_queue<vector<int>> heap;
        
        for (int i = 0; i < n; ++i) {
            heap.push(vector<int>{-matrix[i][0], i, 0});
        }
        vector<int> ptr;
        while (k) {
            ptr = heap.top(); heap.pop();
            if (ptr[2] < n - 1) {
                heap.push(vector<int>{-matrix[ptr[1]][ptr[2] + 1], ptr[1], ptr[2] + 1});
            }
            k--;
        }
        return -ptr[0];
    }
};
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