# 378 Kth Smallest Element in a Sorted Matrix

## 378 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.

Solution 1 : Heap
Here is the step of my solution:

1. Build a minHeap of elements from the first row.
2. Do the following operations k-1 times :
Every time when you poll out the root(Top Element in Heap), you need to know the row number and column number of that element(so we can create a tuple class here), replace that root with the next element from the same column.

After you finish this problem, thinks more :

1. For this question, you can also build a min Heap from the first column, and do the similar operations as above.(Replace the root with the next element from the same row)
2. What is more, this problem is exact the same with Leetcode373 Find K Pairs with Smallest Sums, I use the same code which beats 96.42%, after you solve this problem, you can check with this link:
https://discuss.leetcode.com/topic/52953/share-my-solution-which-beat-96-42
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;
}
}

Solution 2 : Binary Search
The key point for any binary search is to figure out the "Search Space". For me, I think there are two kind of "Search Space" -- index and range(the range from the smallest number to the biggest number). Most usually, when the array is sorted in one direction, we can use index as "search space", when the array is unsorted and we are going to find a specific number, we can use "range".

Let me give you two examples of these two "search space"

1. index -- A bunch of examples -- https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/ ( the array is sorted)
2. range -- https://leetcode.com/problems/find-the-duplicate-number/ (Unsorted Array)

The reason why we did not use index as "search space" for this problem is the matrix is sorted in two directions, we can not find a linear way to map the number and its index.

public class Solution {
public int kthSmallest(int[][] matrix, int k) {
int lo = matrix[0][0], hi = matrix[matrix.length - 1][matrix[0].length - 1] + 1;//[lo, hi)
while(lo < hi) {
int mid = lo + (hi - lo) / 2;
int count = 0,  j = matrix[0].length - 1;
for(int i = 0; i < matrix.length; i++) {
while(j >= 0 && matrix[i][j] > mid) j--;
count += (j + 1);
}
if(count < k) lo = mid + 1;
else hi = mid;
}
return lo;
}
}

Java Alternative Solution 2 with line-by-line comment
The following solution inspired by solution 2 might be a bit more verbose, but I found it consistent with my basic binary search template and easier to understand. This might not be a perfect solution, but this could help u to understand the idea behind the ingenious solution provided by the original Author

public int kthSmallest(int[][] matrix, int k) {
// num of rows and cols in matrix
int rows = matrix.length, cols = matrix[0].length;
// get the lowest and highest possible num, will shrink search space according to the two nums
// [lo, hi] is our initial search range
int lo = matrix[0][0], hi = matrix[rows - 1][cols - 1] ;
while(lo <= hi) {
int mid = lo + (hi - lo) / 2;
int count = 0,  maxNum = lo;
// for each row, we r going to find # of nums < mid in that row
for (int r = 0, c = cols - 1; r < rows; r++) {
while (c >= 0 && matrix[r][c] > mid) c--;   // this row's c has to be smaller than the c found in last row due to the sorted property of the matrix
if (c >= 0) {
count += (c + 1); // count of nums <= mid in matrix
maxNum = Math.max(maxNum, matrix[r][c]); // mid might be value not in matrix, we need to record the actually max num;
}
}
if (count == k) return maxNum;
else if (count < k) lo = mid + 1;
else hi = mid - 1;
}
// 1) Q: Why we return lo at the end:
//    A: Here lo=hi+1, for hi, we found <k elems, for lo, we found >=k elem, lo must have duplicates in matrix, return lo
// 2) Q: Why lo exist in the matrix
//    A: for lo which is only 1 more than hi, we could find some extra nums in matrix so that there r >=k elems, so lo it self must exist in the matrix to meet the requirement
return lo;
}

Actually, if u've done problem 719. Find K-th Smallest Pair Distance
The idea behind the two problems with binary search is exactly the same:
Similar solution for 719. Find K-th Smallest Pair Distance

Also don't forget to check out this fantastic post by @fun4LeetCode
Approach the problem using the "trial and error" algorithm