0541MediumArrayBinary SearchSortingHeap (Priority Queue)Matrix

Kth Smallest Drift Reading

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Trains the technique from

LeetCode 378Kth Smallest Element in a Sorted Matrix

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A calibration bench records the drift of a sensor at n oven temperatures and n supply voltages, giving a square table matrix of n rows and n columns. matrix[i][j] is the drift in micro-units, and it may be negative when the sensor reads low. Raising the temperature never lowers the drift and neither does raising the voltage, so every row of matrix reads left to right in non-decreasing order and every column reads top to bottom in non-decreasing order.

Given k, return the k-th smallest drift value in the table. Repeated values each take their own place in that ordering, so a table whose entries are all equal answers with that value for every k.

Aim for a routine that does not need room for all n * n readings at once.

Examples

Example 1

Input: matrix = [[2, 6, 8], [4, 7, 10], [9, 12, 14]], k = 5
Output: 8

Listed in order the nine readings run 2, 4, 6, 7, 8, 9, 10, 12, 14. The fifth entry of that list is 8.

Example 2

Input: matrix = [[5, 5], [5, 9]], k = 3
Output: 5

The four readings in order are 5, 5, 5, 9. The third of them is 5, because each repeat occupies its own place.

Example 3

Input: matrix = [[2, 5, 9], [3, 6, 10], [4, 7, 11]], k = 4
Output: 5

In order the readings run 2, 3, 4, 5, 6, 7, 9, 10, 11, and the fourth is 5.

Constraints

n == matrix.length == matrix[i].length
1 <= n <= 300
-10^9 <= matrix[i][j] <= 10^9
Every row and every column of matrix is in non-decreasing order.
1 <= k <= n^2

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def kth_smallest(matrix: list[list[int]], k: int) -> int:

Java

public int kthSmallest(int[][] matrix, int k)

The intended approach

Time O(n * log(hi - lo)) where hi and lo are the largest and smallest entries; each candidate costs one O(n) staircase walk.Space O(1) beyond the input table.

Search over the range of drift values rather than over table positions. The smallest entry sits at the top-left and the largest at the bottom-right, so those bound the search. For a candidate value, count how many entries do not exceed it by starting at the top-right corner and sliding the column pointer left while the current entry is too big, then moving down a row; the pointer never moves right again, so the whole count costs one pass across and one pass down. Keep the smallest candidate whose count reaches `k`; that candidate is itself an entry of the table, so the loop settles on the answer.

Pythonentry point: kth_smallest

def kth_smallest(matrix, k):
    # Binary search over the reading values, not over positions. For a candidate
    # value the count of readings that do not exceed it is found with a single
    # staircase walk: start at the top-right corner and slide left as the row
    # index grows, because each row's cut-off can only move left.
    n = len(matrix)
    lo = matrix[0][0]
    hi = matrix[n - 1][n - 1]
    while lo < hi:
        mid = lo + (hi - lo) // 2
        col = n - 1
        count = 0
        for row in range(n):
            while col >= 0 and matrix[row][col] > mid:
                col -= 1
            count += col + 1
        if count >= k:
            hi = mid
        else:
            lo = mid + 1
    return lo

The obvious version

Copy every entry into one flat list, sort it, and read off position `k - 1`. This ignores the row and column ordering entirely.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_kth_smallest(matrix, k):
    pool = []
    for row in matrix:
        for v in row:
            pool.append(v)
    pool.sort()
    return pool[k - 1]

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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