Solution Approach

The solution implements a binary search on the value range combined with an efficient counting mechanism.

Main Binary Search Logic:

1. Initialize the search range with left = matrix[0][0] (minimum value) and right = matrix[n-1][n-1] (maximum value).

2. While left < right:

   • Calculate the middle value: mid = (left + right) >> 1 (using bit shift for integer division)
   • Use the check function to count how many elements are <= mid
   • If count >= k, it means the kth smallest is at most mid, so we update right = mid
   • Otherwise, the kth smallest is greater than mid, so we update left = mid + 1

3. When the loop ends, left equals right and contains our answer.

The Check Function Implementation:

The check function counts elements <= mid using a staircase search pattern:

def check(matrix, mid, k, n):
    count = 0
    i, j = n - 1, 0  # Start from bottom-left corner
    while i >= 0 and j < n:
        if matrix[i][j] <= mid:
            count += i + 1  # All elements above in this column are also <= mid
            j += 1          # Move right to next column
        else:
            i -= 1          # Current element too large, move up
    return count >= k

The movement pattern forms a staircase from bottom-left to top-right:

• When we find matrix[i][j] <= mid, we know all elements from matrix[0][j] to matrix[i][j] are also <= mid (due to column sorting), so we add i + 1 to our count
• We then move right to explore the next column
• If matrix[i][j] > mid, we move up to find smaller elements

Time and Space Complexity:

• Time: O(n * log(max - min)) where n is the matrix dimension. The binary search runs log(max - min) iterations, and each iteration takes O(n) for counting.
• Space: O(1) - only using a few variables, meeting the requirement of better than O(n²) space complexity.

The algorithm guarantees convergence to an actual matrix element because we're searching for the smallest value where at least k elements are <= to it, and this value must exist in the matrix as the kth smallest element.

Example Walkthrough

Let's walk through finding the 4th smallest element in this matrix:

matrix = [[1,  5,  9],
          [10, 11, 13],
          [12, 14, 15]]
k = 4

Step 1: Initialize Binary Search Range

• left = matrix[0][0] = 1 (smallest element)
• right = matrix[2][2] = 15 (largest element)

Step 2: First Binary Search Iteration

• mid = (1 + 15) >> 1 = 8
• Count elements ≤ 8 using the check function:
  • Start at bottom-left: matrix[2][0] = 12
  • 12 > 8, move up to matrix[1][0] = 10
  • 10 > 8, move up to matrix[0][0] = 1
  • 1 ≤ 8, count += 1 (row 0, col 0), move right to matrix[0][1] = 5
  • 5 ≤ 8, count += 1 (row 0, col 1), move right to matrix[0][2] = 9
  • 9 > 8, done. Total count = 2
• Since count (2) < k (4), update left = 8 + 1 = 9

Step 3: Second Binary Search Iteration

• Range is now [9, 15], mid = (9 + 15) >> 1 = 12
• Count elements ≤ 12:
  • Start at matrix[2][0] = 12
  • 12 ≤ 12, count += 3 (all of column 0), move right to matrix[2][1] = 14
  • 14 > 12, move up to matrix[1][1] = 11
  • 11 ≤ 12, count += 2 (rows 0-1 of column 1), move right to matrix[1][2] = 13
  • 13 > 12, move up to matrix[0][2] = 9
  • 9 ≤ 12, count += 1 (row 0, col 2), move right (out of bounds)
  • Total count = 3 + 2 + 1 = 6
• Since count (6) ≥ k (4), update right = 12

Step 4: Third Binary Search Iteration

• Range is now [9, 12], mid = (9 + 12) >> 1 = 10
• Count elements ≤ 10:
  • Start at matrix[2][0] = 12
  • 12 > 10, move up to matrix[1][0] = 10
  • 10 ≤ 10, count += 2 (rows 0-1 of column 0), move right to matrix[1][1] = 11
  • 11 > 10, move up to matrix[0][1] = 5
  • 5 ≤ 10, count += 1 (row 0, col 1), move right to matrix[0][2] = 9
  • 9 ≤ 10, count += 1 (row 0, col 2), move right (out of bounds)
  • Total count = 2 + 1 + 1 = 4
• Since count (4) ≥ k (4), update right = 10

Step 5: Fourth Binary Search Iteration

• Range is now [9, 10], mid = (9 + 10) >> 1 = 9
• Count elements ≤ 9:
  • Start at matrix[2][0] = 12
  • 12 > 9, move up to matrix[1][0] = 10
  • 10 > 9, move up to matrix[0][0] = 1
  • 1 ≤ 9, count += 1, move right to matrix[0][1] = 5
  • 5 ≤ 9, count += 1, move right to matrix[0][2] = 9
  • 9 ≤ 9, count += 1, move right (out of bounds)
  • Total count = 3
• Since count (3) < k (4), update left = 9 + 1 = 10

Step 6: Convergence

• Now left = right = 10
• The 4th smallest element is 10

The sorted elements would be [1, 5, 9, 10, 11, 12, 13, 14, 15], and indeed the 4th element is 10.

Time and Space Complexity

Time Complexity: O(n * log(max - min))

The algorithm uses binary search on the value range [matrix[0][0], matrix[n-1][n-1]]. The binary search runs for log(max - min) iterations where max is the largest element and min is the smallest element in the matrix.

For each binary search iteration, the check function is called which takes O(n) time. The check function uses a two-pointer technique starting from the bottom-left corner of the matrix, moving either up or right. In the worst case, it traverses at most 2n cells (moving n steps up and n steps right), which is O(n).

Therefore, the overall time complexity is O(n * log(max - min)).

Space Complexity: O(1)

The algorithm only uses a constant amount of extra space for variables like left, right, mid, count, i, and j. No additional data structures are created that depend on the input size. The recursive calls are not used, so there's no additional stack space consumption.

Therefore, the space complexity is O(1).

Common Pitfalls

Pitfall 1: Using Wrong Binary Search Template Variant

The Problem: Using while left < right with right = mid instead of the standard template with while left <= right and right = mid - 1.

Wrong Implementation:

while left < right:
    mid = (left + right) // 2
    if count_less_equal(mid) >= k:
        right = mid  # WRONG - should be mid - 1
    else:
        left = mid + 1
return left  # WRONG - should track first_true_index

Solution: Use the standard template with first_true_index to track the answer:

first_true_index = -1
while left <= right:
    mid = (left + right) // 2
    if count_less_equal(mid) >= k:
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1
return first_true_index

Pitfall 2: Binary Search on Indices Instead of Values

Attempting to binary search on matrix indices (positions) instead of the value range fails because matrix elements aren't in sorted order when viewed by linear position.

Wrong Implementation:

left, right = 0, n * n - 1
mid_element = matrix[mid // n][mid % n]  # NOT sorted order!

Solution: Always search on the value range [matrix[0][0], matrix[n-1][n-1]], not on positions.

Pitfall 3: Off-by-One Error in Counting

Forgetting that array indices are 0-based leads to incorrect counts.

Wrong Implementation:

if matrix[row][col] <= target:
    count += row  # WRONG - should be row + 1

Solution: Add row + 1 because row index i means there are i + 1 elements (indices 0 to i).

Pitfall 4: Using Strict Equality Instead of >=

Using count == k instead of count >= k misses the answer when there are duplicate values.

Wrong Implementation:

if count_less_equal(mid) == k:  # WRONG - misses duplicates
    return mid

Solution: Use count >= k to find the smallest value where at least k elements are <= to it.

Pitfall 5: Starting from Wrong Corner in Count Function

Starting from top-left or bottom-right corner makes it impossible to efficiently count elements.

Wrong Implementation:

row, col = 0, 0  # WRONG - can't decide direction efficiently

Solution: Start from bottom-left (n-1, 0) or top-right (0, n-1) corner where you can make definitive decisions about which direction to move.