378. Kth Smallest Element in a Sorted Matrix

1. 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).

2. Example

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

3. 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

4. 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.

5. Solutions

n = matrix.size()
Time complexity: O(nlog(max(matrix) - min(matrix)))
Space complexity: O(1)

// both row and column are ascending
// while merge sort only use row's feature
class Solution {
    int count_smaller_numbers(const vector<vector<int>> &matrix, int number) {
        const int rows = matrix.size();
        const int columns = matrix.front().size();

        int result = 0;
        for (int i = rows - 1, j = 0; i >= 0 && j < columns;) {
            matrix[i][j] < number ? result += (i + 1), ++j : --i;
        }

        return result;
    }

public:
    int kthSmallest(const vector<vector<int>> &matrix, int k) {
        int result = matrix.front().front();
        for (int left = matrix.front().front(), right = matrix.back().back(); left < right;) {
            int mid = (left + right) / 2;

            count_smaller_numbers(matrix, mid) < k ? result = mid, left = mid + 1 : right = mid;
        }

        return result;
    }
};
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