Find the Kth Smallest Sum of a Matrix With Sorted Rows

This problem can be tackled using a min-heap. By applying a min-heap, we can generate sums in a specific order and ensure that we find the kth smallest sum.

Here’s how you can approach the problem:

Initialize a Min-Heap: You need to keep track of the sum and the indices of the elements contributing to the sum.
Insert the First Element: Start by considering the smallest element from each row and insert the sum along with the indices into the heap.
Extract Min and Generate Next Sums: Repeat the following steps k times:
- Remove the smallest sum from the heap.
- Increment one of the indices contributing to the smallest sum and calculate the new sum.
- Insert the new sum along with the updated indices into the heap.
- Make sure to handle duplicates to avoid processing the same combination again.
Return the kth Smallest Sum: The heap will contain the kth smallest sum after k extractions.

Here’s a Python code:

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from heapq import heappop, heappush

class Solution:
    def kthSmallest(self, mat: List[List[int]], k: int) -> int:
        m, n = len(mat), len(mat[0])
        heap = []
        seen = set()

        # Start with the smallest sum (first element from each row)
        first_sum = sum(row[0] for row in mat)
        heap.append((first_sum, [0] * m))
        seen.add(tuple([0] * m))

        for _ in range(k):
            sum_val, indices = heappop(heap)

            # Increment one of the indices and push the new sum into the heap
            for row in range(m):
                new_indices = list(indices)
                new_indices[row] += 1

                # Check for bounds and duplicates
                if new_indices[row] < n and tuple(new_indices) not in seen:
                    new_sum = sum_val - mat[row][indices[row]] + mat[row][new_indices[row]]
                    heappush(heap, (new_sum, new_indices))
                    seen.add(tuple(new_indices))

        return sum_val

This code leverages a min-heap to maintain the order of sums and extract the kth smallest sum efficiently. The time complexity is (O(k \cdot m \log k)), where (m) is the number of rows, and the space complexity is (O(k \cdot m)).

Identifying Problem Isomorphism

“Find the Kth Smallest Sum of a Matrix With Sorted Rows” can be mapped to “Kth Smallest Element in a Sorted Matrix”. Both problems deal with finding a Kth smallest element in a matrix, albeit with different conditions.

In “Kth Smallest Element in a Sorted Matrix”, you are given a matrix where every row and column is sorted in non-decreasing order, and you need to find the Kth smallest element in it.

In “Find the Kth Smallest Sum of a Matrix With Sorted Rows”, the rows of the matrix are sorted, and you are looking for the Kth smallest sum you can obtain by picking one number from each row.

Both problems use heap data structures to keep track of the smallest elements or sums and are related to sorting and matrix traversal.

“Find the Kth Smallest Sum of a Matrix With Sorted Rows” is more complex as it involves finding sums across multiple rows, whereas “Kth Smallest Element in a Sorted Matrix” is simpler as it only requires finding a single element within the matrix.

10 Prerequisite LeetCode Problems

“1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows” combines Heap data structure and Dynamic Programming. Here are 10 problems as good preparation:

215. Kth Largest Element in an Array: Basic heap problem to find the Kth largest element.
378. Kth Smallest Element in a Sorted Matrix: Another basic problem that combines the heap and matrix concepts.
23. Merge k Sorted Lists: Helps in understanding how to merge sorted entities using heap.
347. Top K Frequent Elements: Another problem to understand the application of heap in frequency related problems.
295. Find Median from Data Stream: The problem provides a good understanding of maintaining a data structure with constant-time insert and logarithmic-time extract operations.
239. Sliding Window Maximum: This problem combines the concepts of heap and sliding window.
322. Coin Change: Basic dynamic programming problem to find the minimum number of coins that make a certain amount of change.
96. Unique Binary Search Trees: This is a dynamic programming problem that counts the number of unique BST’s.
64. Minimum Path Sum: A dynamic programming problem that asks for the minimum path sum in a grid.
416. Partition Equal Subset Sum: This problem requires both understanding of dynamic programming and handling of array-based data.

Problem Analysis and Key Insights

What are the key insights from analyzing the problem statement?

Problem Boundary

What is the scope of this problem?

How to establish the boundary of this problem?

Problem Classification

The problem falls under the domain of “Combinatorial Algorithms” and “Data Structures”. It involves working with arrays (matrices) and requires generating combinations.

What Components

Input Matrix (mat): A 2D array where each row is sorted in non-decreasing order.
Integer (k): The rank of the smallest array sum you need to find.
Array Sums: Generated by choosing exactly one element from each row of the given matrix.
Output: The kth smallest sum among all possible array sums.

This is an optimization problem combined with ranking. You have to traverse a combinatorial space as efficiently as possible to find the kth smallest sum. Given the constraints on m, n, and k, it’s clear that brute-force solutions might not be efficient enough, so optimized combinatorial algorithms are likely needed.

The problem also requires a good understanding of data structures like heaps, sets, or priority queues to manage and retrieve the kth smallest sum efficiently. Therefore, it can be further classified under “Optimization in Combinatorial Space with Data Structure Support”.

Distilling the Problem to Its Core Elements

Fundamental Concept

The fundamental concept this problem hinges on is “Combinatorial Enumeration” with an optimization twist. It requires generating combinations of elements from multiple sets (rows in the matrix), and then sorting those combinations based on their sum to find the kth smallest one.

Simplest Description

Imagine you have lists of numbers, and you can pick one number from each list. Add those numbers up. What’s the kth smallest total you can make?

Core Problem

The core problem is finding the kth smallest sum that can be formed by picking exactly one number from each row of a given 2D array (matrix).

Key Components

Enumeration: Generate all possible combinations of numbers by picking one from each row of the 2D array.
Summation: Calculate the sum for each combination.
Ranking: Sort or prioritize these sums to find the kth smallest one.

Minimal Set of Operations

Initialize a priority queue: To keep track of possible sums and their rankings.
Seed the queue: Start with a sum of the smallest elements from each row.
Expand and Insert: Remove the smallest sum from the queue, expand it by replacing one of its components with the next larger number from the same row, and insert the new sums back into the queue.
Track k: Repeat step 3 until we’ve removed k sums from the queue. The kth sum removed is the answer.

This breaks down the core problem into a minimal set of logical steps that can be implemented to solve it efficiently.

Visual Model of the Problem

Visualizing this problem can help in understanding its intricacies. Here’s how you can approach it:

1. Matrix as Multiple Lists:

Imagine each row in the matrix as a separate list. Place these lists horizontally or vertically next to each other.

Row 1: [1, 3, 11]
Row 2: [2, 4, 6]

2. Paths of Choices:

For each row, you have a choice to pick one number. Visualize each choice as a path you can take. Starting from the first row, draw lines to the numbers you could pick in the next row.

  1 -------> 2
  |         |
  |         +------> 4
  |         |
  |         +------> 6
  |
  +--------> 3
            |
            +------> 2
            |
            +------> 4
            |
            +------> 6
  .
  .
  .

3. Summation Points:

Every path you complete (from top row to bottom row) results in a sum. These are your summation points.

Path: 1 -> 2 -> Sum = 3
Path: 1 -> 4 -> Sum = 5
Path: 1 -> 6 -> Sum = 7
Path: 3 -> 2 -> Sum = 5
Path: 3 -> 4 -> Sum = 7
Path: 3 -> 6 -> Sum = 9
...

4. Sorting Sums:

Visualize sorting these sums on a number line, marking the position of each sum.

Number Line: 3--5--7--9--...

5. kth Position:

On this number line, the kth smallest sum is the answer. You can mark it clearly.

By visualizing the problem this way, you can better understand how to navigate the matrix, what choices are available, and how the sum grows as you make specific choices.

Problem Restatement

You have a 2D matrix, each row of which is sorted in increasing order. You’re required to pick one number from each row and sum those numbers together to form an array sum. Your task is to find out what the kth smallest array sum is among all possible combinations.

Requirements:

One number must be selected from each row to form a sum.
You’re looking for the kth smallest sum among all possible sums.

Constraints:

Number of rows (m) and columns (n) can range from 1 to 40.
Each element in the matrix is between 1 and 5000.
The value of k is at least 1 and can go up to the minimum of 200 or m * n.

This exercise helps us understand that our challenge is to navigate through the matrix in such a way that we efficiently find the kth smallest sum of numbers, one from each row.

Abstract Representation of the Problem

You have a collection of sorted lists, each containing integers. Your task is to identify the kth smallest sum that can be created by picking exactly one element from each list.

Structure and Key Elements:

Multiple sorted lists of integers.
A requirement to select exactly one element from each list.
A goal to find the kth smallest sum among all such combinations.

In this abstraction, the focus is on the fundamental operation of selecting elements across sorted lists to form sums, and then determining the rank (kth smallest) of a particular sum. The specific real-world details like matrices and array indices are excluded to emphasize the structural aspects of the problem.

Terminology

Matrix: A two-dimensional array. In this problem, the matrix contains sorted rows of integers.
Non-decreasing Array: An array where elements are in ascending order or remain constant. In this problem, each row of the matrix is a non-decreasing array.
Array Sum: The sum of elements of an array. In this context, you form an array by picking one element from each row of the matrix and then calculate its sum.
kth Smallest Element: The element that would appear in the kth position if you sorted the array. In this problem, you need to find the kth smallest array sum.
Combinations: The different ways to select one element from each row to form an array. The problem requires finding the kth smallest sum among all such combinations.
Constraints: Limits defined for the problem, such as the size of the matrix or the value of k. Adhering to these constraints is crucial for a valid solution.

Understanding these terms is vital as they form the building blocks of both the problem statement and its potential solutions.

Problem Simplification and Explanation

Imagine you have a few stacks of coins, each with different denominations. You’re allowed to take one coin from each stack and put them in your pocket. You then sum up the values of the coins you took. The question is, what’s the kth smallest sum you can get by doing this?

Key Concepts:

Stacks of Coins (Rows in the Matrix): Each row in the matrix is like a stack of coins with different denominations.
One Coin per Stack (One Element from Each Row): You can only take one coin from each stack when you’re creating your pocket of coins.
Sum of Coins (Array Sum): After picking one coin from each stack, you add them together to get the total sum of the coins in your pocket.
kth Smallest Sum (kth Smallest Array Sum): If you were to make different pockets of coins by taking one coin from each stack in every possible way, you’d get a list of total sums. You have to find which total sum would be the kth smallest.

Interaction of Concepts:

You look at each stack (row) one by one and decide which coin (element) to take.
You’re aiming to find the kth smallest sum, so you need to consider all the different sums you can form.
To find the answer, you have to explore the combinations of coins you can take from each stack, sum them up, and find the kth smallest sum.

The stacks of coins, the act of picking one from each, and the summing up are all tied together in answering that kth smallest sum question.

Constraints

Sorted Rows: The rows in the matrix are sorted in non-decreasing order. This can be exploited for efficient searching or traversing through each row. Algorithms like binary search can be effective.
Small Value Ranges: The values in the matrix range from 1 to 5000 and k is up to a minimum of 200 or n*m. The relatively small range can be exploited in counting sort or similar algorithms.
Row and Column Limits: The constraints of 1 <= m, n <= 40 can potentially allow us to use algorithms that may be too slow for larger datasets but work well within these bounds.
Non-Decreasing Rows: This implies that as we move to the next element in a row, the sum will either remain the same or increase. This is valuable for optimization as we don’t need to look back.
kth Smallest: Because we’re looking for a kth smallest sum, we don’t have to generate all possible sums. This allows us to stop early in some searching algorithms.
Matrix Lengths: Knowing that m == mat.length and n == mat.length[i] can simplify the problem because you’re working with a consistent number of choices from each row.
Minimum and Maximum Bounds: The 1 <= mat[i][j] <= 5000 constraint provides a clear boundary, which could be useful for eliminating choices quickly.
Low Minimum Constraint for k: The k value is constrained to be at least 1, which can eliminate the need for checks against empty sets or subsets.

By recognizing these patterns and constraints, we can tailor our approach to be more efficient, possibly using algorithms optimized for sorted data, small ranges, or other specific conditions.

Sorted Rows: The fact that rows are sorted gives us a strong hint that we can use efficient searching algorithms like binary search to navigate each row.
Small Value Range: The constraint on the value range (1-5000) tells us that we don’t need to consider extremely large or small numbers, making counting sort a viable option for some parts of the problem.
Dimension Constraints: With m and n both capped at 40, the problem size remains manageable. Algorithms with higher time complexity may still be practical solutions.
kth Smallest: The need to find just the kth smallest sum, rather than all possible sums, suggests that we can optimize our solution to stop searching once we’ve found this specific element, potentially reducing time complexity.
Fixed Matrix Size: Knowing the dimensions of the matrix in advance allows us to pre-allocate data structures, reducing runtime overhead.
Boundary Values: The constraints for the matrix elements and k give us minimum and maximum bounds, helping to quickly eliminate unfeasible solutions.

The key insights are that we can potentially use searching algorithms and possibly counting-based methods efficiently. The constraints also imply that a brute-force approach could be feasible but not optimal. Thus, it would be wise to look into more efficient algorithms.

Case Analysis

Additional Examples and Test Cases

1. Smallest Dimensions (`TinyMatrix`)

Input: mat = [[1]], k = 1
Output: 1
Analysis: This test case involves the smallest possible dimensions. Since there’s only one element, the output must be that element itself.

2. Multiple Rows, One Element Each (`SingleElementRows`)

Input: mat = [[1], [2], [3]], k = 1
Output: 6
Analysis: When each row has only one element, the sum is predetermined. The output is the sum of all these single elements (1+2+3 = 6).

3. All Elements are the Same (`UniformElements`)

Input: mat = [[1,1,1], [1,1,1], [1,1,1]], k = 4
Output: 3
Analysis: When all elements are the same, any sum will yield the same result. The output is simply the sum of any k combinations.

4. Row Values Disjoint (`DisjointRows`)

Input: mat = [[1, 2], [100, 101]], k = 3
Output: 102
Analysis: When rows have disjoint sets of elements, the smaller elements of later rows will dominate the kth smallest sum.

5. Maximum Elements (`MaxElements`)

Input: mat = [[5000, 5000], [5000, 5000]], k = 1
Output: 10000
Analysis: This case tests the upper bound of the input values. Regardless of k, the sum would be the same since all elements are at their maximum.

6. Varying Row Lengths (`IrregularRows`)

Input: mat = [[1, 5], [2], [3, 4]], k = 4
Output: 8
Analysis: This case explores what happens when rows have different numbers of elements. Here, the 4th smallest sum is 8 (1+3+4).

7. K equals NM (`KequalsNM`)

Input: mat = [[1, 2], [3, 4]], k = 4
Output: 8
Analysis: When k equals the total number of combinations, the kth smallest sum is the sum of the largest elements in each row.

8. Large K (`LargeK`)

Input: mat = [[1,2,3], [4,5,6]], k = 8
Output: 10
Analysis: When k is large but less than nm, it tests the efficiency of the algorithm to compute higher order sums.

Edge Cases

Smallest Dimensions (TinyMatrix) for the lower limit of dimensionality and values.
Maximum Elements (MaxElements) for the upper limit of values.
K equals NM (KequalsNM) and Large K (LargeK) for k at its boundaries within the constraints.

By analyzing these test cases, we cover a wide array of scenarios, providing a comprehensive understanding of the problem’s requirements and constraints.

The key insights from analyzing the different cases are:

Dimensionality Matters: The number of rows and their lengths have a direct impact on the kth smallest sum. Fewer choices in a row simplify the problem, as seen in SingleElementRows.
Value Ranges: The range of values in each row can significantly affect the output. In DisjointRows, the large gap between row values led to a sum dominated by smaller elements of later rows.
Uniformity: When all elements are the same (UniformElements), the kth smallest sum is straightforward. This could be an avenue for optimization.
Edge Values: Cases like MaxElements test the algorithm’s ability to handle the constraints’ upper bounds.
K Boundaries: When k equals the total number of combinations (KequalsNM) or is significantly large (LargeK), it can influence the computational complexity. These cases test the algorithm’s efficiency.
Variability in Row Length: Different row lengths, as seen in IrregularRows, introduce an additional layer of complexity.

These insights inform us about the various aspects of the problem that we need to be careful about while formulating an efficient solution.

Identification of Applicable Theoretical Concepts

Yes, several mathematical and algorithmic concepts can be applied to this problem:

Heap Data Structure: Given the need to find kth smallest sums, a min-heap can efficiently manage this by only storing the smallest k sums at any time.
Dynamic Programming: The problem requires optimizing over multiple choices in multiple rows. Dynamic programming can be used to store and re-use solutions to subproblems.
Combinatorial Analysis: This is a problem of forming combinations. Combinatorial algorithms can be useful for systematically iterating through sets of combinations.
Sorting Algorithms: Rows are already sorted, which is a property that can be exploited for quicker calculations.
Pruning: Given that rows are sorted, we can do branch and bound or pruning. We don’t need to consider sums that are obviously not going to be the kth smallest sum.
Divide and Conquer: This problem can be broken down into smaller problems that are easier to solve. Each of these smaller problems can then be solved independently, possibly in parallel, before their solutions are combined to give the solution to the original problem.
Priority Queues: Similar to heaps but more general, priority queues could manage the kth smallest sums, allowing for quick insertions and deletions.
Greedy Algorithms: In some cases, making the locally optimal choice can lead to a globally optimal solution, although this may not always be true for this specific problem.
Set Theory: The problem boils down to finding specific sets that satisfy a set of conditions, so understanding set operations could help in formulating the algorithm.
Mathematical Bounds: The constraints can be mathematically analyzed to set upper and lower bounds on possible solutions, potentially reducing the search space.

By understanding and applying these concepts, the problem becomes more manageable and allows for more efficient solutions.

Simple Explanation

Imagine you have a few different bowls of jellybeans. Each bowl has jellybeans of different flavors but sorted by taste, from your least favorite to most favorite.

You’re allowed to pick one jellybean from each bowl to make a small bag for yourself. Now, let’s say you want to make sure the bag has the 5th tastiest combination of jellybeans possible.

The problem asks for a similar thing but with numbers in place of jellybeans and their tastiness levels. You have lists of sorted numbers, like the sorted jellybeans in bowls, and you want to find the sum of numbers (one from each list) that would be the kth smallest sum possible.

So, just like you’d pick jellybeans to make a tasty bag, the task is to pick numbers to find a specific sum.

Problem Breakdown and Solution Methodology

Let’s stick with the jellybean metaphor for easy understanding.

Start at the “Lowest Tastiness Level”: Initially, take the least tasty jellybean (i.e., smallest number) from each bowl (i.e., row) and put them in a bag (i.e., a list of sums).
Create “Tasty Combinations”: Now, consider combinations that involve replacing one of the jellybeans in the bag with a tastier one from the same bowl. Calculate the tastiness level (sum) for each new bag and keep track of them.
Prioritize: Sort these new bags by their tastiness level. Imagine having a priority queue that keeps track of the tastiest combinations you can form, always serving you the least tasty bag first.
Iterate: Each time you pull out the least tasty bag, replace one jellybean in it with a tastier one from the same bowl, calculate the new tastiness level, and put it back in the priority queue.
Find the kth Tastiest: Keep doing this until you’ve pulled out ‘k’ bags. The last bag you pull out will have the kth tastiest combination of jellybeans (i.e., the kth smallest sum).

Changes in Parameters:

If you add more bowls (rows), it adds complexity but doesn’t change the approach.
If you add more jellybeans in a bowl (elements in a row), it gives you more options for tasty combinations but might take longer to find the kth tastiest bag.

Example Case:

Let’s consider the first example: mat = [[1,3,11],[2,4,6]], k = 5.

Start with [1,2] with a sum of 3.
Potential next bags: [3,2] (sum 5) or [1,4] (sum 5).
Pull out [1,4] (it’s one of the least), next options: [3,4] (sum 7), [1,6] (sum 7).
You now have pulled out 5 bags: [1,2], [3,2], [1,4], [3,4], [1,6]. The last one has a sum of 7, which is the answer.

By breaking down the problem into these steps, you can systematically find the kth smallest sum among all possible combinations.

Inference of Problem-Solving Approach from the Problem Statement

Here are some key terms and concepts that guide the approach:

Matrix: The problem presents the data in a 2D array, or matrix, form. This dictates that we likely need nested loops or recursive strategies to explore all elements.
Rows Sorted in Non-decreasing Order: This characteristic allows us to make optimizations. If we pick a larger element from a row, we know we won’t get a smaller sum using elements from that row in the future.
kth Smallest Array Sum: The goal is to find a specific order statistic, the kth smallest sum. This tells us we can use techniques like heap-based priority queues to keep track of the smallest sums encountered.
One Element from Each Row: This constraint simplifies the problem because we only need to select one element from each row for consideration. It eliminates the need to consider all possible subsets.
Constraints on ’m’, ’n’, and ‘k’: The given constraints tell us that we can’t have more than 40 rows and 40 elements in each row. This suggests that a computational solution is feasible, but we still need to be mindful of optimization.
Non-decreasing Array: The rows are sorted in non-decreasing order, which means binary search or other optimized searching methods could be utilized for specific parts of the problem.
Array Sum: This term implies that we’ll be performing addition operations, which are computationally cheap and can often be optimized.

Each of these terms and conditions gives us clues on how to approach the problem efficiently. For example, the sorted nature of the rows suggests that we don’t need to sort them ourselves, saving time. The constraints give us a hint about the computational limits we need to adhere to, informing us that while brute-force may be infeasible, optimized computational solutions could work.

How did you infer from the problem statement that this problem can be solved using ?

Simple Explanation of the Proof

I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?

Initialize a Priority Queue: The priority queue will hold potential sums and a state of indices representing which element was picked from each row.
Seed the Queue: Initially populate the priority queue with the sum of the smallest elements from each row. Keep track of the indices.
Process k Elements: Extract the smallest sum from the queue, k times. Each time, generate new sums by replacing one element from each row, then re-insert the new sums into the queue.
Terminate: After k smallest sums have been extracted, the last one is your answer.

Granular Steps

Initialization
- Create a min-heap or priority queue.
- Declare variables to store current state and sum.
Seeding
- Calculate the initial sum of the smallest elements in each row.
- Insert this sum into the queue, along with the corresponding indices.
Iteration
- Pop the smallest sum from the priority queue.
- Increment k.
- For each row, create a new sum by replacing the current row’s element with the next smallest element in that row.
- Insert these new sums back into the queue.
Termination
- Once k elements have been extracted, return the last extracted sum.

Independent Parts

Row Processing: Each row’s smallest element can be found and added to the initial sum independently of other rows.
Priority Queue Operations: Inserting and extracting elements can happen independently of the row processing.

Repeatable Patterns

Queue Processing: The pattern of popping the smallest element, generating new sums, and inserting them back is repeated k times.
Element Replacement: For each row, the action of finding the next smallest element and calculating a new sum is repeated.

By breaking down the problem in this way, we create a roadmap that simplifies implementation and helps us focus on one piece at a time.

Solution Approach and Analysis

Approach to Solving the Problem

Priority Queue Setup: Think of the priority queue as a “magic box” that always gives you the smallest item when you ask for it. Initialize it with the sum of the smallest elements from each row of the matrix.
Seeding the Queue: Imagine you’re at a buffet, and you initially load your plate with the least expensive items. In the same way, calculate the sum of the smallest elements in each row and insert this sum into the priority queue. Also, store the position of each chosen element.
Iterative Sum Generation: Like adding a sprinkle of truffle on your dish at the buffet, swap out one cheap item for a more expensive one and see how the sum changes. For each popped sum, generate new sums by replacing one element in each row with the next smallest element in that row. Insert these back into the priority queue.
Counting Iterations: You are allowed to modify your plate from the buffet k times. After the kth smallest sum has been taken out from the priority queue, that’s your answer.
Return the kth Sum: The kth smallest sum extracted from the priority queue is the final answer.

Operations and Parameter Changes

Matrix Dimensions: If the matrix dimensions are large, the priority queue could get very large, affecting time complexity.
Value of k: If k is close to the maximum possible value (n * m), the solution will need to nearly exhaust the priority queue, increasing time complexity.
Element Values: If the matrix has many repeated values, this can actually make the algorithm faster, as fewer unique sums need to be considered.

Demonstration with an Example

Let’s take Example 1 from the problem statement:

mat = [[1,3,11],[2,4,6]], k = 5

Step 1: Initialize priority queue. Insert initial sum 1+2=3.

Step 2: Store the indices [0,0] to represent that you’ve chosen the 0th element from both rows.

Step 3: Start popping from the queue.

First pop gives 3, next smallest sums are 1+4=5 and 3+2=5. Insert both into the queue.
Second pop gives 5, next smallest sums are 3+4=7 and 1+6=7. Insert both into the queue.
Third pop gives another 5, skip as it is duplicate.
Fourth pop gives 7, next smallest sums are 3+6=9 and 11+2=13. Insert both into the queue.
Fifth pop gives another 7.

Step 4: We’ve popped k=5 sums from the priority queue.

Step 5: The last popped sum is 7, which is our answer.

In this manner, you can solve the problem step by step.

Identify Invariant

The invariant in this problem is the non-decreasing order of the rows in the given matrix. This characteristic holds true throughout the problem and is key to generating the kth smallest array sum efficiently.

The non-decreasing order ensures that as you proceed to find the kth smallest sum, you can generate newer sums in a controlled manner. When you replace an element from a particular row to generate a new sum, you are guaranteed that the newer sum will be greater than or equal to the previous one. This controlled increment allows the priority queue to maintain the smallest sum at the top, helping you reach the kth smallest sum in an efficient way.

Understanding this invariant can guide you in crafting an effective solution, as it ensures that the priority queue always gives you the next smallest array sum, leading you toward the kth smallest array sum.

Identify Loop Invariant

The loop invariant in this problem would depend on the specific implementation of the algorithm. However, one common loop invariant, when using a priority queue to solve this problem, might be that the queue always contains the smallest sums that can be formed by choosing one element from each row of the matrix, up to the current point in the algorithm’s execution.

In simpler terms, if you’re iterating through the possible combinations to find the kth smallest sum, the invariant is that at any point before or after an iteration of the loop, the priority queue will hold the smallest sums that can be generated given the rows processed so far. This ensures that when you pop an element off the queue, you are always working with the smallest sum available at that point in time.

Maintaining this loop invariant is crucial for ensuring that you ultimately find the kth smallest sum accurately and efficiently.

Thought Process

Basic Thought Process and Steps:

Understand the Problem: We’re given an m x n matrix with each row sorted in non-decreasing order. The aim is to find the kth smallest sum formed by choosing exactly one element from each row. This is a combinatorial problem with a sorting twist.
Identify Cues:
- The rows are sorted.
- We need to pick one element from each row.
- We’re interested in finding the kth smallest sum.
- The constraints on m, n, and k suggest that a brute-force solution that would require O(n^m) time is not feasible.
Algorithm Choice: Priority Queue to store sums and process the smallest sums first.
Data Structures Needed:
- Priority Queue (Heap) to store sums
- Tuple to store (sum, list of indices representing elements contributing to the sum)
Key Operations:
- Heap insertion
- Heap removal
Approach:
- Initialize the heap with the smallest possible sum (sum of the first elements from each row) and the corresponding indices.
- Pop the heap to get the smallest current sum and its indices.
- Generate new sums by replacing one element in the current sum, pushing these new sums into the heap.
- Keep a count of how many sums have been popped off the heap. Stop when the kth smallest sum is reached.

Python3 Code:

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import heapq

def kthSmallest(mat, k):
    m, n = len(mat), len(mat[0])
    
    # Start with the smallest sum (sum of the first elements in each row)
    heap = [(sum(row[0] for row in mat), [0]*m)]
    
    # Use a set to keep track of sums we've seen
    seen = set()
    seen.add(tuple([0]*m))
    
    while heap:
        val, indices = heapq.heappop(heap)
        
        # If this is the kth smallest sum, return it
        k -= 1
        if k == 0:
            return val
        
        # Generate the next possible sums by adding the next element in each row one by one
        for i in range(m):
            new_indices = list(indices)
            if new_indices[i] < n - 1:
                new_indices[i] += 1
                
                # Calculate the new sum
                new_val = val - mat[i][indices[i]] + mat[i][new_indices[i]]
                
                # Use a tuple to represent the list of indices, for easy comparison
                tuple_indices = tuple(new_indices)
                
                # If we haven't seen this set of indices before, add to heap and set
                if tuple_indices not in seen:
                    seen.add(tuple_indices)
                    heapq.heappush(heap, (new_val, new_indices))

Insights:

The sorted nature of the rows allows the use of Priority Queue to efficiently traverse through possible sums.
The need to pick one element from each row led to the use of a tuple of indices to represent a sum.
The Priority Queue handles the “kth smallest” requirement by naturally presenting sums in ascending order.

Establishing Preconditions and Postconditions

Parameters:

Inputs:
- mat: a 2D list of integers
- k: an integer
Types:
- mat: List[List[int]]
- k: int
Context:
- mat: Represents the m x n matrix where each row is sorted in non-decreasing order.
- k: Represents the index of the smallest sum to return.

Preconditions:

State of the Program: None, as this is a standalone function.
Constraints:
- m == len(mat)
- n == len(mat[i])
- 1 <= m, n <= 40
- 1 <= mat[i][j] <= 5000
- 1 <= k <= min(200, nm)
Specific State: No specific state requirement.

Method Functionality:

Expectation:
- The method is expected to return the kth smallest array sum formed by choosing one element from each row of mat.
Interaction:
- The method only interacts with its input parameters and does not rely on or alter any global state.

Postconditions:

State: No change in the state of the program or the values of the parameters.
Return Value:
- Returns an integer representing the kth smallest array sum.
Side Effects:
- None, the method is free of side effects.

Error Handling:

Response:
- The function does not explicitly handle errors related to precondition violations.
Exception/Special Value:
- No exception handling is implemented. If preconditions are not met, behavior is undefined.

Problem Decomposition

1. Problem Understanding:

The problem asks us to find the kth smallest sum that can be formed by selecting one element from each row of a given 2D matrix (mat). Each row in mat is sorted in non-decreasing order.

2. Initial Breakdown:

Reading and validating input.
Generating all possible sums.
Sorting these sums.
Finding the kth smallest sum.

Reading and validating input:
- Ensure the matrix and k meet the specified constraints.
Generating all possible sums:
- Enumerate through each row and accumulate sums for each combination.
Sorting these sums:
- Utilize an efficient sorting algorithm to sort the sums.
Finding the kth smallest sum:
- Directly index into the sorted array to retrieve the kth smallest sum.

4. Task Identification:

The task of generating all possible sums involves iterating through each row, which is a repeatable pattern.
The task of sorting these sums is also a generic, reusable task.

5. Task Abstraction:

Generate Sums: Sufficiently abstract to take any 2D matrix and return a list of sums by selecting one element from each row.
Sort Sums: Generic sorting function that sorts a list of integers.

6. Method Naming:

validateInput(mat, k): Validates the inputs.
generateAllSums(mat): Generates all possible sums.
sortSums(sums): Sorts the sums in non-decreasing order.
findKthSmallest(sums, k): Retrieves the kth smallest sum.

7. Subproblem Interactions:

validateInput should be run first to ensure the subsequent methods can operate on the inputs.
generateAllSums should be executed next to get all the possible sums.
After obtaining the sums, sortSums is used to sort them.
Finally, findKthSmallest should be used to get the kth smallest sum from the sorted list.
No significant dependencies except that findKthSmallest relies on the output of sortSums, which in turn relies on generateAllSums.

From Brute Force to Optimal Solution

Brute Force Solution:

Generate all possible sums: Loop through each row and accumulate sums for each combination. This would be O(m^n) where m is the number of elements in each row and n is the number of rows.
Sort the sums: After generating all possible sums, sort them. Sorting would take O(N log N) time where N is the total number of sums.
Retrieve the kth sum: Finally, retrieve the kth sum from the sorted list, which takes O(1) time.

Python Code for Brute-Force

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from itertools import product

def brute_force_kthSmallest(mat, k):
    rows = len(mat)
    if rows == 0:
        return -1

    # Generate all possible sums
    all_sums = []
    for row_elems in product(*mat):
        all_sums.append(sum(row_elems))

    # Sort the sums
    all_sums.sort()

    # Retrieve the kth smallest sum
    return all_sums[k-1]

Inefficiencies:

Time Complexity: The time complexity of generating all possible sums is O(m^n) and sorting them is O(N log N). Thus, the overall time complexity is O(m^n + N log N).
Space Complexity: We are storing all the sums, so the space complexity is O(N).

Optimization:

Heap Method: Instead of generating all possible sums, use a min-heap to keep track of the smallest sums. By using heap push and pop, we can find the kth smallest sum without sorting all sums.
Pruning: Not all sums need to be calculated. We can prune the sums that are surely larger, focusing only on potential kth smallest sums.

Python Code for Optimized Solution

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import heapq

def optimized_kthSmallest(mat, k):
    rows, cols = len(mat), len(mat[0])
    if rows == 0:
        return -1
    
    heap = [(sum(row[0] for row in mat), [0]*rows)]
    
    seen = set()
    seen.add(tuple([0]*rows))
    
    while k > 0:
        val, indices = heapq.heappop(heap)
        k -= 1

        for i in range(rows):
            new_indices = list(indices)
            if new_indices[i] < cols - 1:
                new_indices[i] += 1

                if tuple(new_indices) not in seen:
                    seen.add(tuple(new_indices))
                    heapq.heappush(heap, (val - mat[i][indices[i]] + mat[i][new_indices[i]], new_indices))

    return val

Impact on Time and Space Complexity:

Time Complexity: With the heap, each push/pop operation is O(log k). We might do this up to k * rows times, making it O(k * rows * log k).
Space Complexity: The heap takes O(k) space, and the set seen also takes O(k). Hence, the overall space complexity is O(k).

This optimized solution significantly reduces both time and space complexity.

Code Explanation and Design Decisions

1. Initial Parameters:

mat: It’s a 2D matrix where each row is sorted. This matrix holds all the rows for which we need to find the kth smallest sum of elements.
k: It signifies the kth smallest sum we are interested in.

These are the main driving factors for our problem.

2. Primary Loop:

The primary loop involves iterating through the min-heap, heap. Each iteration pops the smallest sum and its corresponding row indices from the heap and pushes new sums by adding the next element from each row.

Each iteration represents the extraction of the smallest unprocessed sum and the possibility of new, smaller sums coming into consideration.

3. Conditions or Branches:

The main condition is if new_indices[i] < cols - 1. This checks whether we have reached the end of any row while calculating sums.

Another condition is if tuple(new_indices) not in seen. This avoids adding duplicate sums to the heap.

These conditions are critical to ensure that we only consider valid and unique sums, respecting the problem’s constraints.

4. Updates or Modifications:

new_indices[i] += 1: We update the index to consider the next element in the ith row. This reflects new possible sums that could be smaller and candidates for kth smallest sum.
heapq.heappush(): We push the new sum and its indices back into the heap. This is essential for maintaining a pool of smallest sums.

These modifications are critical to expand our search for potential kth smallest sums while keeping computational resources in check.

5. Invariant:

The invariant is that the heap always contains sums that are candidates for being the kth smallest sum, sorted in increasing order. The seen set ensures that these sums are unique.

This invariant ensures that we can safely take the smallest element from the heap at each step and be assured it’s the next smallest sum we haven’t considered.

6. Final Output:

The final output is the kth smallest sum. It satisfies the problem’s requirement by considering all possible sums from the elements of the rows in the given matrix and efficiently determining which one is the kth smallest.

Coding Constructs

1. High-Level Strategies:

Heap Data Structure: Min-heap to efficiently keep track of the smallest sums.
Memoization: Use of a set to remember already computed sums to avoid duplicates.

2. Purpose to a Non-Programmer:

The code finds the kth smallest sum possible by picking one number from each row in a table of numbers, where each row is sorted from small to big.

3. Logical Elements:

Looping: To iterate through possibilities.
Conditions: To filter out invalid or duplicate sums.
Sorting: Indirectly through the use of a min-heap.
Memoization: To store already seen sums.

4. Algorithm in Plain English:

Start by adding the smallest possible sum to a priority list.
Remove the smallest sum from the list, and generate new sums by replacing one element in the original sum with the next bigger element in the same row.
Keep track of sums you’ve already seen.
Repeat this process until you’ve found the kth smallest sum.

5. Key Steps:

Initialize the heap with the smallest sum (sum of the first elements in each row).
Pop the heap to get the smallest current sum and generate new sums by advancing in each row.
Push new sums into the heap if they haven’t been processed before.
Repeat until we’ve found the kth smallest sum.

6. Algorithmic Patterns:

Priority Queue: For efficiently identifying the next smallest sum.
Breadth-First Search: In essence, it explores all possible sums in order of their size, somewhat similar to breadth-first traversal in graphs.
Memoization: To avoid recalculating the same sum.

Language Agnostic Coding Drills

Your mission is to deconstruct this code into the smallest possible learning units, each corresponding to a separate coding concept. Consider these concepts as unique coding drills that can be individually implemented and later assembled into the final solution.

Dissect the code and identify each distinct concept it contains. Remember, this process should be language-agnostic and generally applicable to most modern programming languages.
Once you’ve identified these coding concepts or drills, list them out in order of increasing difficulty. Provide a brief description of each concept and why it is classified at its particular difficulty level.
Next, describe the problem-solving approach that would lead from the problem statement to the final solution. Think about how each of these coding drills contributes to the overall solution. Elucidate the step-by-step process involved in using these drills to solve the problem. Please refrain from writing any actual code; we’re focusing on understanding the process and strategy.

Targeted Drills in Python

1. General Concepts:

Looping: Iterating through a list

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# Simple for-loop to iterate through a list and print elements
nums = [1, 2, 3, 4]
for num in nums:
    print(num)

Conditions: Using `if` statements

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# If condition to filter out specific numbers
x = 5
if x > 3:
    print("x is greater than 3")

Sorting: Using Min-Heap

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import heapq

# Initialize a heap and add elements
heap = []
heapq.heappush(heap, 3)
heapq.heappush(heap, 1)
heapq.heappush(heap, 4)

# Remove and return smallest element
print(heapq.heappop(heap))

Memoization: Using a set to store unique elements

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# Initialize a set
seen = set()

# Add an element
seen.add(2)

# Check for an element
if 2 not in seen:
    print("Element not found")

2. Problem-Specific Concepts:

Combining elements to make sums

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# Make sums of elements of two lists and store in a heap
list1 = [1, 7]
list2 = [3, 5]
sum_heap = []

for elem1 in list1:
    for elem2 in list2:
        heapq.heappush(sum_heap, elem1 + elem2)

Why essential?: Our main problem involves generating sums by combining elements from multiple lists. Understanding how to generate these sums is crucial.

3. Assembling the Final Solution:

Initialize the Min-Heap: First, you use Python’s heapq library to initialize a heap data structure. You start by pushing the smallest possible sum into this heap.
Looping & Heap Operations: You enter a loop that will run until you find the kth smallest sum. Inside this loop, you pop the smallest sum off the heap, which uses the Sorting concept.
Generating New Sums: Using the popped sum, you generate new sums (problem-specific concept) by advancing through each row.
Check Conditions & Memoization: Before pushing new sums into the heap, you check whether you’ve seen this sum before using a set for Memoization. If not, you push it back into the heap.

By following these steps and using each of the drilled concepts, you can assemble them into a cohesive solution for the initial problem.

Q&A

What are the reasons for making these mistakes in the given code?