You are given an n x n binary grid board. In each move, you can swap any two rows with each other, or any two columns with each other.
Return the minimum number of moves to transform the board into a chessboard board. If the task is impossible, return -1.
A chessboard board is a board where no 0's and no 1's are 4-directionally adjacent.
Example 1:
Input: board = [[0,1,1,0],[0,1,1,0],[1,0,0,1],[1,0,0,1]] Output: 2 Explanation: One potential sequence of moves is shown. The first move swaps the first and second column. The second move swaps the second and third row.
Example 2:
Input: board = [[0,1],[1,0]] Output: 0 Explanation: Also note that the board with 0 in the top left corner, is also a valid chessboard.
Example 3:
Input: board = [[1,0],[1,0]] Output: -1 Explanation: No matter what sequence of moves you make, you cannot end with a valid chessboard.
Constraints:
n == board.lengthn == board[i].length2 <= n <= 30board[i][j] is either 0 or 1.This approach involves matching the row and column patterns against a valid chessboard configuration and calculating the minimum number of swaps needed. The idea is to compare the current board's rows and columns to the two possible valid chessboard row and column patterns.
The function movesToChessboard attempts to transform the given board into a chessboard. It determines if the transformation is possible by verifying the row and column patterns. The pattern matching checks if the lines have the correct number of 1s or 0s to resemble a chessboard line pattern. movesToTransform calculates the minimum swaps needed to align each row and column pattern to either of two potential chessboard patterns.
Java
Time Complexity: O(n2) due to examining each cell multiple times for pattern matching.
Space Complexity: O(n) for storing intermediate lists.
Recursive Approach with Memoization:
This approach involves breaking the problem down into smaller subproblems, solving each using recursion, and storing the results of these subproblems to avoid redundant calculations. This is a classic dynamic programming approach that trades time complexity for space complexity by using a cache to remember previously computed results.
Memoization helps in optimizing the recursive calls by saving the results of subproblems, making future calls faster and reducing time complexity significantly.
This Python solution makes use of a dictionary, memo, to store computed Fibonacci numbers. The base cases are n = 0 and n = 1 where the function returns the number itself. For other cases, the result is calculated recursively and stored in the memo dictionary.
C
Time Complexity: O(n)
Space Complexity: O(n) due to the recursion call stack and memoization dictionary.
Iterative Approach with Tabulation:
This approach uses an iterative method to fill up a table (array), which stores previously computed values of the subproblems (in this case, Fibonacci numbers). Tabulation is another form of dynamic programming where you solve the problem iteratively and build the solution bottom-up.
This approach optimizes space usage by using only an array of size n or even reducing it to constant space by utilizing only the last two computed values.
This JavaScript solution calculates Fibonacci numbers iteratively, keeping only the last two values at any given time (fib1 and fib2), thus optimizing space usage significantly, achieving O(1) space complexity while maintaining O(n) time complexity.
Java
Time Complexity: O(n)
Space Complexity: O(1)
| Approach | Complexity |
|---|---|
| Pattern Matching and Position Swapping | Time Complexity: O(n2) due to examining each cell multiple times for pattern matching. |
| Recursive Approach with Memoization | Time Complexity: O(n) |
| Iterative Approach with Tabulation | Time Complexity: O(n) |
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