Kth Smallest Element in a Sorted Matrix LeetCode Solution

problem – Kth Smallest Element in a Sorted Matrix

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.

Kth Smallest Element in a Sorted Matrix LeetCode Solution in Java

public class Solution {
    public int kthSmallest(int[][] matrix, int k) {
        int n = matrix.length;
        PriorityQueue<Tuple> pq = new PriorityQueue<Tuple>();
        for(int j = 0; j <= n-1; j++) pq.offer(new Tuple(0, j, matrix[0][j]));
        for(int i = 0; i < k-1; i++) {
            Tuple t = pq.poll();
            if(t.x == n-1) continue;
            pq.offer(new Tuple(t.x+1, t.y, matrix[t.x+1][t.y]));
        }
        return pq.poll().val;
    }
}

class Tuple implements Comparable<Tuple> {
    int x, y, val;
    public Tuple (int x, int y, int val) {
        this.x = x;
        this.y = y;
        this.val = val;
    }
    
    @Override
    public int compareTo (Tuple that) {
        return this.val - that.val;
    }
}

Kth Smallest Element in a Sorted Matrix LeetCode Solution in C++

class Solution
{
public:
	int kthSmallest(vector<vector<int>>& matrix, int k)
	{
		int n = matrix.size();
		int le = matrix[0][0], ri = matrix[n - 1][n - 1];
		int mid = 0;
		while (le < ri)
		{
			mid = le + (ri-le)/2;
			int num = 0;
			for (int i = 0; i < n; i++)
			{
				int pos = upper_bound(matrix[i].begin(), matrix[i].end(), mid) - matrix[i].begin();
				num += pos;
			}
			if (num < k)
			{
				le = mid + 1;
			}
			else
			{
				ri = mid;
			}
		}
		return le;
	}
};

Kth Smallest Element in a Sorted Matrix LeetCode Solution in Python

class Solution(object):
    def kthSmallest(self, matrix, k):
        lo, hi = matrix[0][0], matrix[-1][-1]
        while lo<hi:
            mid = (lo+hi)//2
            if sum(bisect.bisect_right(row, mid) for row in matrix) < k:
                lo = mid+1
            else:
                hi = mid
        return lo

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