Leetcode: Kth Smallest Element in a Sorted Matrix

Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix.

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

Example:

matrix = [
   [ 1,  5,  9],
   [10, 11, 13],
   [12, 13, 15]
],
k = 8,

return 13.
Note: 
You may assume k is always valid, 1 ≤ k ≤ n2.

 1 public class Solution {
 2     public int kthSmallest(int[][] matrix, int k) {
 3         Comparator<Tuple> comp = new Comparator<Tuple>() {
 4             public int compare(Tuple tp1, Tuple tp2) {
 5                 return tp1.val - tp2.val;
 6             }
 7         };
 8         
 9         PriorityQueue<Tuple> minHeap = new PriorityQueue<Tuple>(matrix.length, comp);
10         
11         int res = 0;
12         
13         for (int row=0; row<matrix.length; row++) {
14             minHeap.offer(new Tuple(row, 0, matrix[row][0]));
15         }
16         
17         
18         for (int i=1; i<=k; i++) {
19             Tuple tp = minHeap.poll();
20             if (i == k) {
21                 res = tp.val;
22                 break;
23             }
24             if (tp.y < matrix.length-1) minHeap.offer(new Tuple(tp.x, tp.y+1, matrix[tp.x][tp.y+1]));
25         }
26         return res;
27     }
28     
29     public class Tuple {
30             int x;
31             int y;
32             int val;
33             public Tuple(int i, int j, int value) {
34                 this.x = i;
35                 this.y = j;
36                 this.val = value;
37             }
38     }
39 }

 1 public class Solution {
 2     public int kthSmallest(int[][] matrix, int k) {
 3         int lo = matrix[0][0], hi = matrix[matrix.length - 1][matrix[0].length - 1] + 1;//[lo, hi)
 4         while(lo < hi) {
 5             int mid = lo + (hi - lo) / 2;
 6             int count = 0,  j = matrix[0].length - 1;
 7             for(int i = 0; i < matrix.length; i++) {
 8                 while(j >= 0 && matrix[i][j] > mid) j--;
 9                 count += (j + 1);
10             }
11             if(count < k) lo = mid + 1;
12             else hi = mid;
13         }
14         return lo;
15     }
16 }