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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.
1public int movesToChessboard(int[][] board) {
2 int n = board.length;
3 // Helper method
4 int getTransformMoves(int[] line) {
5 int count = 0, n = line.length;
6 for (int i = 0; i < n; i++) {
7 if (line[i] == i % 2) count++;
8 }
9 return count;
10 }
11 ...The Java solution follows similar logic, examining each row and column against ideal chessboard patterns and computes the necessary swaps. Validate the initial pattern of rows and checking against possible transformations directly reflect 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.
1def fibonacci(n, memo={}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 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.
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.