Transform to Chessboard

Identifying Problem Isomorphism

“Flip Game II” is approximately isomorphic to “Transform to Chessboard”.

In “Transform to Chessboard”, you’re given a 2D array, and the task is to convert this board into a chessboard by swapping any two rows or any two columns. The number of swaps should be minimized.

In “Flip Game II”, you are given a string representation of a game board and can flip two consecutive “++” into “–”. The goal is to determine if the first mover will win if both players play optimally.

The problems are isomorphic in the sense that they both require a transformation of a game board with an aim towards an optimal solution. The tasks differ, but the underlying thinking required for both revolves around board manipulation, considering several moves ahead, and attempting to minimize or optimize a particular outcome.

“Transform to Chessboard” is more challenging due to its additional layer of complexity - needing to minimize the number of swaps, which requires a higher level of strategic decision-making.

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class Solution:
    def movesToChessboard(self, board):
        board_size = len(board)

        if board_size <= 1:
            return 0

        row_representation = [''.join(str(cell) for cell in row) for row in board]
        row_counter = collections.Counter(row_representation)
        row_keys = list(row_counter)

        if len(row_keys) != 2 or abs(row_counter[row_keys[0]] - row_counter[row_keys[1]]) > 1 or abs(row_keys[0].count('1') - row_keys[0].count('0')) > 1 or any(a == b for a, b in zip(*row_keys)):
            return -1

        row_diff = sum(board[0][i] != (i % 2) for i in range(board_size))
        col_diff = sum(board[i][0] != (i % 2) for i in range(board_size))
        row_diff = board_size - row_diff if row_diff % 2 != 0 or (board_size % 2 == 0 and (board_size - row_diff) < row_diff) else row_diff
        col_diff = board_size - col_diff if col_diff % 2 != 0 or (board_size % 2 == 0 and (board_size - col_diff) < col_diff) else col_diff

        return (row_diff + col_diff) // 2

Problem Classification

This problem deals with a chessboard represented as a 2D matrix and operations on the matrix.

Components:

Input: An n x n binary grid board represented as a list of lists. Here, n is the size of the board, which ranges from 2 to 30.
Operations: In each move, we are allowed to swap any two rows with each other or any two columns with each other.
Output: We need to return the minimum number of moves to transform the board into a chessboard board. If the task is impossible, we should return -1.

This is a Combinatorial Search/ Optimization problem. We need to find the optimal sequence of operations (swapping rows or columns) that minimizes the number of moves needed to reach the desired state, which is a chessboard-like configuration. If there is no such sequence, we should return -1 indicating the task is impossible.

Difficulty: It’s an intermediate to hard level problem considering the constraints and the combinatorial nature of the problem.
Relevance: It can be related to real-world problems involving the optimization of tasks with constraints.
Knowledge Required: The problem requires a good understanding of matrix manipulation and search/optimization algorithms.

Thought Process

The problem asks to transform a binary grid board into a chessboard pattern where no 0’s and 1’s are 4-directionally adjacent. The only operations we can perform are row swaps and column swaps.

The cues in the problem statement are:

A grid and operations on the grid (swapping rows or columns).
The final configuration of the grid should resemble a chessboard.

Based on these cues, we can follow these steps to devise a solution:

For a grid to be transformed into a chessboard pattern, it must satisfy a particular condition: each row or column must have an equal number of 0s and 1s (or differ by at most one if the size of the grid is odd). Therefore, we first check if the grid satisfies this condition. If not, we can directly return -1, as transforming the board into a chessboard will be impossible.
If the grid passes the first condition, we then check if the rows and columns of the grid can be arranged to form a chessboard pattern. For this, we find out all possible chessboard patterns (with 0 starting or with 1 starting) and check if the given grid can be transformed into one of them. If it can be, we then count the minimum number of swaps needed to transform the current grid into a valid chessboard pattern.
Finally, return the minimum number of swaps needed. If the grid cannot be transformed into any valid chessboard pattern, return -1.

Language Agnostic Coding Drills

Dissecting the code into distinct concepts:
- Variable Initialization and Assignment: This is one of the basic concepts of any programming language. Variables are used to store data, which can be used throughout the code.
- Basic Data Structures: The code involves the use of a list, which is a fundamental data structure. Knowledge of how to initialize, access, and manipulate lists is required.
- Conditional Statements: The if statements are used to control the flow of the code based on certain conditions.
- Loops: The for loops are used for iterating over elements of a list. Understanding how to use loops is an essential part of coding.
- String Manipulation: Strings are being manipulated for creating a row representation. This involves converting integers to strings and joining them.
- Built-in Functions and Methods: Functions like sum(), len(), list(), range(), and any() are used. Understanding these built-in functions and methods is important for coding.
- Collections - Counter: It’s a dictionary subclass for counting hashable objects. In this case, it’s used to count the frequency of row representations.
- Comprehensions: They provide a concise way to create new lists. In the code, list comprehensions are used to generate new lists, like row_representation, row_diff and col_diff.
- Zip Function: The zip() function is used to combine two lists in a way that the i-th element of the first list is paired with the i-th element of the second list.
- Lambda Expressions: Anonymous function defined with lambda. The use of lambda function is crucial in any() to determine the condition.
Order of Increasing Difficulty:
- Variable Initialization and Assignment: This is the simplest concept as it’s the most fundamental part of coding.
- Basic Data Structures: The concept of lists isn’t very difficult, but it requires understanding of more advanced concepts than just variable initialization and assignment.
- Conditional Statements: These involve logical thinking, so they’re a step above basic concepts like variables and data structures.
- Loops: Looping requires understanding of control flow, which makes them a bit more complex.
- String Manipulation: This concept requires an understanding of data types and built-in functions.
- Built-in Functions and Methods: Requires knowledge of Python’s in-built libraries, and understanding of various types of data structures and their methods.
- Collections - Counter: It’s a more advanced topic, as it’s part of Python’s collections module, and requires understanding of dictionaries and hashing.
- Comprehensions: Comprehensions require an understanding of both loops and conditional statements, and thus, they are more advanced.
- Zip Function: This function requires knowledge of tuples and iterable unpacking, so it’s more difficult than the previous concepts.
- Lambda Expressions: This is one of the more advanced topics, as it requires a good understanding of functions and Python’s syntax for anonymous functions.
Problem-Solving Approach:
- Initialize the variables and determine the size of the board. This will be used in the upcoming steps.
- Convert the rows of the board into strings. This step prepares data for further processing.
- Use the collections.Counter to count the occurrence of each unique row. This information will be used to check whether the board can be converted into a chessboard.
- Check the conditions to determine if it’s possible to convert the board into a chessboard. If not, return -1 immediately.
- Calculate the number of differences for rows and columns. This involves looping through the board and comparing each cell with the expected cell in a chessboard.
- Update the differences for rows and columns. If the number of differences is odd or the difference is less than half of the board size, replace the difference with the board size minus the difference.
- Return the total number of swaps required to convert the board into a chessboard. This involves summing up the differences for rows and columns and dividing by 2.

Each of the identified coding concepts is crucial in the overall solution. The problem-solving approach is a logical sequence of these coding concepts, where each step contributes to reaching the solution.

Targeted Drills in Python

Coding Drills for Identified Concepts:

Variable Initialization and Assignment:

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a = 10
b = 20
print(a, b)

Basic Data Structures:

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my_list = [1, 2, 3, 4, 5]
print(my_list)

Conditional Statements:

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a = 10
if a > 5:
    print("a is greater than 5")
else:
    print("a is not greater than 5")

Loops:

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for i in range(5):
    print(i)

String Manipulation:

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str_num = str(5)
print(str_num)

Built-in Functions and Methods:

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my_list = [1, 2, 3, 4, 5]
print(len(my_list))

Collections - Counter:

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from collections import Counter
my_list = ['a', 'b', 'c', 'a', 'b', 'a']
print(Counter(my_list))

Comprehensions:

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my_list = [i for i in range(5)]
print(my_list)

Zip Function:

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list1 = [1, 2, 3]
list2 = ['a', 'b', 'c']
print(list(zip(list1, list2)))

Lambda Expressions:

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square = lambda x: x**2
print(square(5))

Problem-Specific Coding Drills:

Converting a list of integers to a string:

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my_list = [1, 0, 1, 0]
str_list = ''.join(str(i) for i in my_list)
print(str_list)

This drill is important as the problem requires a string representation of each row to compare rows.

Counting the occurrences of each unique row:

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from collections import Counter
rows = ['1010', '0101', '1010', '0101']
print(Counter(rows))

This is crucial as the problem involves determining the frequency of unique rows.

Integrating Coding Drills to solve the initial problem:

All the coding drills can be assembled together in the context of the initial problem, the steps are as follows:

Start with initializing the variables such as board size, row representation, and counters for different rows, and differences for rows and columns.
Create a string representation of each row using a list comprehension.
Use collections.Counter to count the frequency of each unique row.
Apply conditional statements to determine whether the board can be converted into a chessboard based on the count of unique rows.
If it’s possible to convert the board, calculate the number of differences for each row and column using loops and built-in functions.
Update the differences using conditional statements and arithmetic operations.
Finally, return the total number of swaps required to convert the board into a chessboard.

These individual coding drills can be combined following the problem-solving approach defined in the previous steps to provide a cohesive solution to the problem.

Q&A

An observation is that, in a valid ChessBoard, any rectangle inside the board with corner NW, NE, SW, SE (NW here means north-west) must satisfy (NW xor NE) == (SW xor SE), where xor is exclusive or. Here we call it validness property.

Since the swap process does not break this property, for a given board to be valid, this property must hold. A corollary is that given a row, any other row must be identical to it or be its inverse. For example if there is a row 01010011 in the board, any other row must be either 01010011 or 10101100.

With this observation, we only need to consider the first column when we’re swapping rows to match the ChessBoard condition. That is, it suffies to find the minimum row swap to make the first column be 010101…^T or 101010…^T. (here ^T means transpose.)

Similarly, it suffies to consider the first row when swapping columns.

Now the problem becomes solvable, with the following steps:

Check if the given board satisfy the validness property defined above. Find minimum row swap to make the first column valid. If not possible, return -1. Find minimum column swap to make the first row valid. If not possible, return -1. Return the sum of step 2 and 3.

Great, you have provided an in-depth analysis of the problem, and you’re right on track. As you mentioned, checking for the validness property is crucial and then finding the minimum swaps is the core part of the solution.

In this solution, the code checks if each row in the board satisfies the validness property (by comparing each row to the first one and checking the number of 1s). It also checks the number of 1s in the first column and row. After these checks, it calculates the number of swaps for rows and columns needed to make the first column and row valid. If the size of the board is even or the parity of the number of swaps and the size of the board is the same, it returns the sum of the minimum number of swaps for rows and columns. Otherwise, it returns -1.

a board where no 0s and no 1s are 4-directionally adjacent?

What does it mean?

Is 0 1 0 0 1 1 0 1 1 0 0 1 0 0 1 a valid solution?

Is 0 1 0 0 0 1 0 1 1 1 0 1 0 0 0 a valid solution?

In the context of this problem, “4-directionally adjacent” means any cell that can be reached from the current cell by moving up, down, left, or right. It does not include diagonally adjacent cells.

For a board to be a valid chessboard, no 0’s and no 1’s should be 4-directionally adjacent. This means that if we take any cell on the board, all its 4-directional neighbours should have the opposite value.

In a valid chessboard, you will see alternating 1’s and 0’s in each row and column. For example:

are valid chessboards. Notice how in each of these, no two 1’s are adjacent and no two 0’s are adjacent in a 4-directional sense.

For your examples:

0 1 0 0 1
1 0 1 1 0
0 1 0 0 1

is not a valid chessboard because, for example, the two 0’s in the top row are 4-directionally adjacent.

0 1 0 0 0
1 0 1 1 1
0 1 0 0 0

is also not a valid chessboard because, for example, the three 0’s in the top row are 4-directionally adjacent.