Kth Smallest in a Sorted Matrix

1. Step 1: Initialize Solution.kthSmallest

   Replace the empty starter with the first real state owned by Solution.kthSmallest. A small, named state is easier to verify than a complete algorithm. Establish it before adding the branch or loop that changes it.

2. Step 2: Pass the First case

   Complete the readable core algorithm for one representative Interview case. Treat rows as sorted sequences and advance only the row whose head was popped.

3. Step 3: Harden the Duplicates boundary

   Repair the reviewed boundary and pass the complete submission contract. The heap contains the smallest unconsumed value from every row. Its minimum is therefore the smallest unconsumed matrix value globally; after k pops the last value is exactly kth smallest.

• Do not deduplicate equal values; each matrix position counts toward k.
• python specific rapid fire
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• LeetCode 378 — Kth Smallest Element in a Sorted Matrix

• Update a Bounded Heap(opens in a new tab)

  Update a Bounded Heap isolates after each value, the heap contains the largest min(k, processed_count) values seen so far. That focused state discipline is required when implementing kth smallest sorted matrix as a complete Interview Problem.

Push the first value of each row with row and column indices. Repeat k times: pop the smallest and push the next value from that same row.

Why it works: The heap contains the smallest unconsumed value from every row. Its minimum is therefore the smallest unconsumed matrix value globally; after k pops the last value is exactly kth smallest.

class Solution:
    def kthSmallest(self, matrix, k):
        """
        Checkpoint 1: initialize the state owned by this Interview contract.
        Checkpoint 2: assemble the primary transition without hiding the boundary.
        """
        import heapq
        heap = [(row[0], index, 0) for index, row in enumerate(matrix)]
        heapq.heapify(heap)
        value = None
        for _ in range(k):
            value, row, column = heapq.heappop(heap)
            if column + 1 < len(matrix[row]):
                heapq.heappush(heap, (matrix[row][column + 1], row, column + 1))
        return value

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