First Completely Painted Row or Column

To solve this problem, we need to keep track of the number of cells painted in each row and each column.

For each index i in the array arr, we can compute the position (r, c) in the matrix mat where the integer arr[i] appears. After painting this cell, we’ll increment the number of painted cells in row r and in column c. If any row or column becomes fully painted, we return the index i.

Here are the steps:

Create an array rows of size m initialized with zeros to keep track of the number of painted cells in each row.
Create an array cols of size n initialized with zeros to keep track of the number of painted cells in each column.
Create a dictionary positions where for each value v in the matrix mat, we store its position (r, c).
For each index i in arr, update the rows and cols array using the position from the positions dictionary. If any row or column is fully painted, return i.

Here’s the code:

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class Solution:
    def firstCompleteIndex(self, arr: List[int], mat: List[List[int]]) -> int:
        m, n = len(mat), len(mat[0])
        rows = [0] * m
        cols = [0] * n
        positions = {}

        # Create a dictionary of positions for each value in mat
        for i in range(m):
            for j in range(n):
                positions[mat[i][j]] = (i, j)

        for i in range(len(arr)):
            r, c = positions[arr[i]]
            rows[r] += 1
            cols[c] += 1

            if rows[r] == n or cols[c] == m:
                return i

        return -1  # This line is technically unnecessary given problem constraints, but included for completeness.

For each index in arr, we get the position of the value in mat using the positions dictionary, then update the rows and cols arrays. If we find a fully painted row or column, we return the current index.

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def firstCompleteIndex(arr: List[int], mat: List[List[int]]) -> int:
    m = len(mat)
    n = len(mat[0])
    map = {}
    for i in range(m):
        for j in range(n):
            map[mat[i][j]] = [i, j]
    row = [0] * m
    col = [0] * n
    for i in range(len(arr)):
        x = map[arr[i]]
        row[x[0]] += 1
        col[x[1]] += 1
        if row[x[0]] == n or col[x[1]] == m:
            return i
    return -1

Problem Classification

The problem falls under the domain of Arrays and Matrix manipulation in the field of Computer Science and Algorithms.

What:

Array: An input array ‘arr’ containing all integers from 1 to m*n.
Matrix: An input matrix ‘mat’ of dimensions mn, also containing all integers from 1 to mn.
Painting Process: A process where we sequentially go through each index of the array, and “paint” the corresponding cell in the matrix.
Completely Painted Row/Column: A state where all cells in a row or a column of the matrix have been painted.
Smallest Index: The output which is the smallest index in the array ‘arr’ where a row or column in the matrix is completely painted.

This is an Iterative Search problem. It involves iterating over the array and searching for the smallest index at which a certain condition (a row or column being completely painted) is met.

The problem involves manipulating data structures (an array and a matrix), which suggests the general domain. It asks for the smallest index at which a row or column is painted, implying that we need to iterate over the array and check the state of the matrix after each step, which classifies it as an Iterative Search problem.

This problem falls under the domain of Array and Matrix Processing in computer science and algorithm design. The problem’s main components are:

Input Data: We have an integer array arr and an integer matrix mat as inputs. Both contain all integers from 1 to m * n.
Data Transformation: The task is to iterate through each index in the array arr, use the integer at that index to locate a corresponding cell in the matrix mat, and “paint” it (probably marking it in some way).
End Condition: The process continues until we reach a point where a whole row or column of the matrix mat is painted.
Output: We need to return the smallest index i in arr at which this happens.

This problem can be further classified as a simulation problem, where we simulate the painting process according to the given rules and return the smallest index where the process completes for either a row or a column.

The categorization is based on the tasks we need to perform, the nature of the data structures we’re working with (arrays and matrices), and the simulation aspect of the problem. The solution will likely involve array/matrix manipulation and keeping track of the painted cells until a complete row or column is painted.

Language Agnostic Coding Drills

Concept Identification:
a. Array and Matrix Manipulation: The very basic concept involved in this code is the manipulation of arrays and matrices, which includes accessing their elements and understanding how they are indexed.
b. Use of Dictionaries: The code employs a dictionary to map each integer in the matrix to its position in the matrix.
c. Looping over an array: The code iterates over the array arr, using each item to access the corresponding cell in the matrix and “paint” it.
d. Tracking the progress: The code uses two additional arrays to track the number of painted cells in each row and column of the matrix.
e. Exit condition check: At each step, the code checks if a whole row or column has been painted. This is a slightly more complex concept, as it involves keeping track of the state of multiple data structures and evaluating an exit condition.
Difficulty Ordering:
a. Array and Matrix Manipulation: This is a basic concept in most programming languages.
b. Looping over an array: This builds upon the concept of array manipulation, introducing the notion of iteration.
c. Use of Dictionaries: Although this is not a complex concept, it might be more difficult for beginners to understand compared to simple array manipulations and iterations.
d. Tracking the progress: This involves understanding and updating the state of multiple arrays concurrently, which might be slightly more challenging for someone who is just starting out with programming.
e. Exit condition check: This is the most difficult concept, as it involves checking the state of multiple data structures and determining when a certain condition is met.
Problem-solving approach:

The first step in solving the problem is to create a map of the matrix that allows easy access to the positions of the integers. Next, the code creates two arrays to keep track of the number of painted cells in each row and column of the matrix. Then, it iterates over the array arr, using the map to locate the corresponding cell in the matrix, and “paints” it by incrementing the count of painted cells in the corresponding row and column. At each step, it checks if any row or column has been completely painted. If it finds one, it immediately returns the current index in arr. If no row or column is completely painted after iterating over all of arr, it returns -1.

Targeted Drills in Python

Coding Drills for Identified Concepts:

a. Array and Matrix Manipulation:

 ```python
 # Creating an array and a matrix
 arr = [1, 2, 3]
 mat = [[1, 2], [3, 4]]

 # Accessing elements
 print(arr[0]) # Output: 1
 print(mat[1][0]) # Output: 3
 ```

b. Looping over an array:

 ```python
 arr = [1, 2, 3]
 for i in arr:
     print(i) # Output: 1, then 2, then 3
 ```

c. Use of Dictionaries:

 ```python
 map = {1: 'one', 2: 'two', 3: 'three'}

 # Accessing values using keys
 print(map[1]) # Output: 'one'
 ```

d. Tracking the progress:

 ```python
 progress = [0, 0, 0]

 for i in range(3):
     progress[i] += 1
     print(progress) # Output: [1, 0, 0], then [1, 1, 0], then [1, 1, 1]
 ```

e. Exit condition check:

 ```python
 progress = [0, 0, 0]

 for i in range(3):
     progress[i] += 1
     if progress[i] == 1:
         print(f"Exit condition met at index {i}") # Output: Exit condition met at index 0, 1, 2
         break
 ```

Problem-Specific Concepts and Drills

The problem-specific concept in this problem is the “painting” of the cells in the matrix and tracking the progress. Here’s a simplified drill for that:

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mat = [[1, 2], [3, 4]]
arr = [1, 2, 3, 4]

# Mapping the cells in the matrix
map = {}
for i in range(len(mat)):
    for j in range(len(mat[0])):
        map[mat[i][j]] = [i, j]

# Painting the cells
row_progress = [0, 0]
col_progress = [0, 0]

for i in range(len(arr)):
    x = map[arr[i]]
    row_progress[x[0]] += 1
    col_progress[x[1]] += 1
    print(f"Cell {arr[i]} painted at index {i}, Row Progress: {row_progress}, Column Progress: {col_progress}")

Integration
To solve the original problem, we need to integrate all these concepts in the following order:
a. Begin by creating the map for the cells in the matrix using the concept from the dictionary drill.
b. Initialize arrays to track the row and column progress using the concept from the array manipulation drill.
c. Loop over the array arr using the looping concept. In each iteration:
```
 - "Paint" the corresponding cell in the matrix and update the progress arrays using the concepts from the progress tracking and matrix manipulation drills.

 - Check if a whole row or column has been painted using the concept from the exit condition check drill. If so, return the current index.
```
d. If the loop completes without finding a completely painted row or column, return -1.