What is the time complexity of Find the Kth Smallest Sum of a Matrix With Sorted Rows?

The optimal solution for Find the Kth Smallest Sum of a Matrix With Sorted Rows runs in O(k log m) time and uses O(m) space. This complexity arises because we perform k operations on a heap that can grow to size m, where m is the number of rows.

What is the brute force solution for Find the Kth Smallest Sum of a Matrix With Sorted Rows?

The brute force approach for Find the Kth Smallest Sum of a Matrix With Sorted Rows is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach generates all possible sums by selecting one element from each row. It then sorts these sums to find the k-th smallest. This method is straightforward but inefficient for larger matrices. The optimal Optimal Solution approach improves this to O(k log m).

What algorithm or data structure is used to solve Find the Kth Smallest Sum of a Matrix With Sorted Rows?

Find the Kth Smallest Sum of a Matrix With Sorted Rows uses the following concepts: Array, Binary Search, Heap (Priority Queue), Matrix. The recommended patterns to study are: Heap (Priority Queue), Array.

What are the interview tips for Find the Kth Smallest Sum of a Matrix With Sorted Rows?

Always clarify the problem constraints and edge cases before diving into coding. Think about how to optimize your approach from brute force to a more efficient solution. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Find the Kth Smallest Sum of a Matrix With Sorted Rows?

Forgetting to handle the indices properly when generating new sums. Not using a priority queue effectively, leading to incorrect results.

#1439

Find the Kth Smallest Sum of a Matrix With Sorted Rows

Hard

ArrayBinary SearchHeap (Priority Queue)MatrixHeap (Priority Queue)Array

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(k log m)
Space	O(1)	O(m)

💡

Intuition

Time O(k log m)Space O(m)

Using a min-heap allows us to efficiently track the smallest sums without generating all combinations. We can incrementally build the sums and explore new candidates based on the current smallest sum.

⚙️

Algorithm

3 steps

1Step 1: Initialize a min-heap and add the smallest sum (the first element of each row) along with their indices.
2Step 2: Pop the smallest sum from the heap and push new sums formed by incrementing the indices of the rows.
3Step 3: Repeat until we pop the k-th smallest sum.

solution.py17 lines

1# Full working Python code
2import heapq
3
4def kthSmallest(mat, k):
5    m, n = len(mat), len(mat[0])
6    min_heap = []
7    initial_sum = sum(row[0] for row in mat)
8    heapq.heappush(min_heap, (initial_sum, [0] * m))
9    for _ in range(k):
10        current_sum, indices = heapq.heappop(min_heap)
11        for i in range(m):
12            if indices[i] + 1 < n:
13                new_indices = indices[:]
14                new_indices[i] += 1
15                new_sum = current_sum - mat[i][indices[i]] + mat[i][new_indices[i]]
16                heapq.heappush(min_heap, (new_sum, new_indices))
17    return current_sum

ℹ

Complexity note: This complexity arises because we perform k operations on a heap that can grow to size m, where m is the number of rows.

1Using a min-heap allows us to efficiently track the smallest sums without generating all combinations.
2The matrix's sorted rows mean that we can always find the next smallest sum by incrementing indices.

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