Jeremy Fisher - 4 months ago 67

Python Question

I know this has been asked before but this question is about my specific code. I am trying to do a psuedo QuickSelect algorithm where I compare k to the midpoint of a sub interval of a sorted matrix.

I keep getting a timeout error.

Here is the matrix:

`matrix = [`

[ 1, 5, 9],

[10, 11, 13],

[12, 13, 15]

],

k = 8

Here is my code:

`def kthSmallest(self, matrix, k):`

"""

:type matrix: List[List[int]]

:type k: int

:rtype: int

"""

return self.matrixRecur(matrix, (0, 0), (len(matrix) - 1, len(matrix[0]) - 1), k)

def matrixRecur(self, splicedMatrix, left, right, k):

start_row, start_col = left

end_row, end_col = right

mid_row = (start_row + end_row)/2

mid_col = (start_col + end_col)/2

i = start_row

j = start_col

lcount = 0

while(not (i == mid_row and j == mid_col)):

if j < len(splicedMatrix[0]):

j += 1

else:

j = 0

i += 1

lcount += 1

if k == lcount:

return splicedMatrix[mid_row][mid_col]

elif k < lcount:

return self.matrixRecur(splicedMatrix, (start_row, start_col), (mid_row, mid_col), k)

else:

return self.matrixRecur(splicedMatrix, (mid_row, mid_col + 1), (end_row, end_col), k-lcount)

I pass in tuples to

`matrixRecur`

`(row, col)`

`(0, 0)`

`(n, n)`

`matrixRecur`

`k`

`k`

`lcount`

`lcount`

`k`

`lcount`

I am running this on an interview question site and the site continues to tell me my code times out. I am not seeing why.

Answer

Your above approach will not work. As your matrix is row and column wise sorted. Consider the matrix like below

```
10, 20, 30, 40
15, 25, 35, 45
24, 29, 37, 48
32, 33, 39, 50
```

Your approach will fail in this case. You are getting time out because you are traversing the whole 2D matrix. Worst case time complexity would be `O(mn)`

(m and n are respectively number of rows and columns).

You can use a **min-heap** to solve this problem.

Algorithm:

```
1. Build a min heap of first row elements. A heap entry also stores row number and column number.
2. for(int i=0;i<k;i++)
Get minimum element (or root) from min heap.
Find row number and column number of the minimum element.
Replace root with the next element from same column and min-heapify the root.
3. Return the last extracted root.
```

Time complexity : `O(n + kLogn)`

Source : Kth smallest in 2D matrix

Source (Stackoverflow)

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