Sponsored
Sponsored
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.
Time Complexity: O(n2) due to examining each cell multiple times for pattern matching.
Space Complexity: O(n) for storing intermediate lists.
1def movesToChessboard(board):
2 n = len(board)
3
4 def movesToTransform(line):
5 ones = sum(line)
6 if not (n // 2 <= ones <= (n + 1) // 2):
7 return float('inf') # Not possible to transform to a line
8
9 line1Count = sum(line[i] == i % 2 for i in range(n))
10 line2Count = sum(line[i] != i % 2 for i in range(n))
11
12 if n % 2:
13 # Odd size board, should have a dominant line pattern
14 if line1Count % 2:
15 return n - line2Count
16 return n - line1Count
17 return min(n - line1Count, n - line2Count)
18
19 # Check if transformation possible
20 firstRow = board[0]
21 if any(set(board[i]) != set(firstRow) for i in range(n)):
22 return -1
23
24 firstColumn = [board[i][0] for i in range(n)]
25 if any(set([board[i][j] for i in range(n)]) != set(firstColumn) for j in range(n)):
26 return -1
27
28 rowMoves = movesToTransform([board[i] for i in range(n)])
29 colMoves = movesToTransform([board[i][0] for i in range(n)])
30 return (rowMoves + colMoves) // 2The 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.
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.
Time Complexity: O(n)
Space Complexity: O(n) due to the recursion call stack and memoization dictionary.
1#include <stdio.h>
2
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.
Time Complexity: O(n)
Space Complexity: O(1)
1public class Fibonacci {
2 This C program uses a global array memo with a fixed size to store results of subproblems. Only values of Fibonacci numbers are calculated if they haven't been computed before, leveraging memoization to achieve efficiency.
The Java approach uses an array fib to store Fibonacci numbers calculated in sequence. The for loop computes values from the bottom-up, storing each result to be used in subsequent calculations.