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# 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`

• 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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