Question

378. Kth Smallest Element in a Sorted Matrix

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Description

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.

Solutions

Solution 1

class Solution:
  def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
    def check(matrix, mid, k, n):
      count = 0
      i, j = n - 1, 0
      while i >= 0 and j < n:
        if matrix[i][j] <= mid:
          count += i + 1
          j += 1
        else:
          i -= 1
      return count >= k

    n = len(matrix)
    left, right = matrix[0][0], matrix[n - 1][n - 1]
    while left < right:
      mid = (left + right) >> 1
      if check(matrix, mid, k, n):
        right = mid
      else:
        left = mid + 1
    return left

class Solution {
  public int kthSmallest(int[][] matrix, int k) {
    int n = matrix.length;
    int left = matrix[0][0], right = matrix[n - 1][n - 1];
    while (left < right) {
      int mid = (left + right) >>> 1;
      if (check(matrix, mid, k, n)) {
        right = mid;
      } else {
        left = mid + 1;
      }
    }
    return left;
  }

  private boolean check(int[][] matrix, int mid, int k, int n) {
    int count = 0;
    int i = n - 1, j = 0;
    while (i >= 0 && j < n) {
      if (matrix[i][j] <= mid) {
        count += (i + 1);
        ++j;
      } else {
        --i;
      }
    }
    return count >= k;
  }
}

class Solution {
public:
  int kthSmallest(vector<vector<int>>& matrix, int k) {
    int n = matrix.size();
    int left = matrix[0][0], right = matrix[n - 1][n - 1];
    while (left < right) {
      int mid = left + right >> 1;
      if (check(matrix, mid, k, n)) {
        right = mid;
      } else {
        left = mid + 1;
      }
    }
    return left;
  }

private:
  bool check(vector<vector<int>>& matrix, int mid, int k, int n) {
    int count = 0;
    int i = n - 1, j = 0;
    while (i >= 0 && j < n) {
      if (matrix[i][j] <= mid) {
        count += (i + 1);
        ++j;
      } else {
        --i;
      }
    }
    return count >= k;
  }
};

func kthSmallest(matrix [][]int, k int) int {
  n := len(matrix)
  left, right := matrix[0][0], matrix[n-1][n-1]
  for left < right {
    mid := (left + right) >> 1
    if check(matrix, mid, k, n) {
      right = mid
    } else {
      left = mid + 1
    }
  }
  return left
}

func check(matrix [][]int, mid, k, n int) bool {
  count := 0
  i, j := n-1, 0
  for i >= 0 && j < n {
    if matrix[i][j] <= mid {
      count += (i + 1)
      j++
    } else {
      i--
    }
  }
  return count >= k
}