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 ≤ n^2.

Thoughts:

1. Heap:
   1. Build a minHeap of elements from the first row.
   2. Repeat for k - 1 times: poll out the current element from the queue, get its row and col number. if row is not at the end (n-1), push a new element at the same column, next row into the queue
   3. return the kth smallest value by pop out of the queue

2. Binary Search:

   1. binary search the value in matrix to get its rank on the matrix:

   2. if rank < k: low = mid + 1

   3. hi = mid

   4. return lo(or hi) in the end

Code: Heap T: O(k*logn); S: O(n)

class Solution(object):
    def kthSmallest(self, matrix, k):
        """
        :type matrix: List[List[int]]
        :type k: int
        :rtype: int
        """
        n = len(matrix)
        pq = []
        for j in range(n):
            heapq.heappush(pq, Element(0, j, matrix[0][j]))

        for i in range(k - 1):
            e = heapq.heappop(pq)
            if e.row == n - 1: continue
            heapq.heappush(pq, Element(e.row + 1, e.col, matrix[e.row + 1][e.col]))

        return heapq.heappop(pq).val

class Element(object):
    def __init__(self, row, col, val):
        self.row = row
        self.col = col
        self.val = val

    def __lt__(self, other):
            return self.val < other.val

    def __eq__(self, other):
            return self.val == other.val

Code: Binary Search: T:O(n*log(max - min));S: O(1)

class Solution {
    public int kthSmallest(int[][] matrix, int k) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0;
        int m = matrix.length, n = matrix[0].length;
        int l = matrix[0][0], h = matrix[m - 1][n-1];
        while(l < h){
            int mid = l + (h - l >> 1);
            int cnt = 0, j = n - 1;
            // rank mid value in matrix
            for(int i = 0; i < m; i++){
                while(j >=0 && matrix[i][j] > mid) j --; // since column is ascending
                cnt+= (j + 1);                    // so matrix[i + 1][j] >= matrix[i][j] -> do not have to
                                                        // reset j here
            }

            if(cnt < k) l = mid + 1;
            else h = mid;
        }

        return l; // or h
    }
}

Good Reference: https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/85173/Share-my-thoughts-and-Clean-Java-Code