Minimum Swaps to Arrange a Binary Grid in Java

Swapping of rows to arrange a binary grid with minimum swaps is an exciting problem that requires the conversion of the given binary grid to a particular form. The objective is to make sure that for every row i in the grid, it has atleast n-1-i trailing zeroes, where n is the total number of rows in the grid, this problem can be solved efficiently with greedy algorithm, BFS and Segment tree.

Example:

Input

Output:

 
3

Explanation

For row 0, it needs at least 2 trailing zeros. The row with 2 trailing zeros is at index 2. We swap it up to index 0 with two swaps. For row 1, it needs at least 1 trailing zero. The row with 1 trailing zero is now at index 2. We swap it up to index 1 with one swap. For row 2, it needs at least 0 trailing zeros, which it already has, so no swap is needed. The total number of swaps is 3.

Approach 1: BFS (Breadth-First Search)

The Breadth-First Search (BFS) approach treats the problem as a state space search where each state represents a configuration of the grid, and we aim to find the shortest path (minimum swaps). It explores all possible configurations of the grid by swapping rows and uses a queue to track the configurations to explore.

Algorithm

Step 1: Create an array trailingZeros to store the number of trailing zeros for each row. Use a queue to store configurations of the grid. Use a set to track visited configurations to avoid redundant processing.

Step 2: Calculate Trailing Zeros: Iterate through each row of the grid and count the number of trailing zeros. Store these counts in the trailingZeros array.

Step 3: BFS Setup: Enqueue the initial configuration of trailingZeros and mark it as visited. Initialize a variable: swaps to count the number of swaps needed.

Step 4: BFS Loop: Dequeue a configuration. Check if the current configuration is valid. If valid, return the number of swaps taken to reach this configuration.

Step 5: Generate all possible configurations by swapping adjacent rows. Enqueue each new configuration if it has not been visited and mark it as visited.

Step 6: If the queue is exhausted without finding a valid configuration, return -1.

Filename: MinimumSwapsBinaryGridBFS.java

Output:

 
3

Time Complexity

In the worst case, BFS might need to explore all possible permutations of rows, which is n!. For each permutation, checking validity takes O(n), and generating new configurations takes O(n⋅n)=O(n²). Therefore, the total time complexity for BFS exploration is O((n!)* n²).

Space Complexity

The queue can contain up to n! configurations, and each configuration is an array of size n. The visited set also stores up to n! configurations, each of size n. Therefore, the space complexity is O((n!)* n).

Approach 2: Using Segment Tree

In this problem, we can use a segment tree to manage the counts of trailing zeros in each row of the grid, allowing us to quickly determine the minimum number of swaps needed to arrange the grid as required.

Algorithm

Step 1: Create an array trailingZeros of size n to store the count of trailing zeros for each row.

Step 2: Counting Trailing Zeros: Loop through each row i from 0 to n-1. For each row, iterate through its elements from right to left (from n-1 to 0). Count the number of consecutive zeros from the end of the row and store this count in trailingZeros[i].

Step 3: Build the Segment Tree Segment Tree Construction: Initialize a segment tree where each leaf node corresponds to a count of trailing zeros. Build the segment tree using the trailingZeros array.

Step 4: Range Queries: For each row i, determine the required number of trailing zeros (n - 1 - i). Use the segment tree to perform range minimum queries to find the minimum trailing zeros in the range from i to the end of the array.

Step 5: If the minimum value found is less than the required zeros, find the corresponding row index.

Step 6: Swap the rows to move the row with the required trailing zeros to the correct position. Update the segment tree to reflect the new trailing zero counts after the swap.

Filename: MinimumSwapsBinaryGridST.java

Output:

 
3

Time Complexity

We iterate through each element of the grid to count trailing zeros, resulting in O(n²). Each query and update operation on the segment tree takes O(logn), and we perform these operations n times hence it is O(nlogn). Therefore, the overall time complexity is O(n² + nlogn).

Space Complexity

Space for Trailing Zeros Array is O(n): We use an array trailingZeros to store the count of trailing zeros for each row. Hence, the space complexity is O(n).

Approach 3: Fenwick Tree (Binary Indexed Tree)

In the context of arranging a binary grid with the minimum number of swaps, a Fenwick Tree can be employed to efficiently keep track of the positions of the rows with enough trailing zeros, this allows us to quickly determine the smallest number of swaps needed to move a row into its correct position.

Algorithm

Step 1: Initialize Variables: n, trailingZeros and FenwickTree: a data structure to keep track of counts and perform efficient prefix sum queries and updates.

Step 2: For each row, count the number of trailing zeros (0s) from right to left. Store the count in the trailingZeros array.

Step 3: Create a Fenwick Tree of size n. Update the tree with the counts of trailing zeros for each row.

Step 4: For each position i from 0 to n-1: Calculate the required number of trailing zeros for the row to be placed at position i (which is n - 1 - i).

Step 5: Use the Fenwick Tree to find the closest row with sufficient trailing zeros. If no such row is found, return -1 (indicating it's not possible to arrange the grid).

Step 6: For each row found, perform the necessary swaps to move the row to the correct position.

Step 7: Update the Fenwick Tree after each swap to reflect the current state of the trailingZeros array. Return the total number of swaps required to arrange the grid.

Filename: MinimumSwapsBinaryGridFT.java

Output:

 
3

Time Complexity

Counting Trailing Zeros: O(n²) for iterating through each element of the grid. Fenwick Tree Operations: O(nlogn) for updates and queries. Overall Time Complexity: O(n² + nlogn).

Space Complexity

The space complexity for Trailing Zeros Array is O(n): We store the number of trailing zeros for each of the n rows. Fenwick Tree also requires O(n) space: It has a size proportional to the number of rows. Therefore, the overall space complexity is O(n).

Approach 4: Greedy Approach Using Trailing Zeros

The problem of arranging a binary grid such that each row i has at least n-1-i trailing zeros can be efficiently solved using a greedy approach, this approach involves counting trailing zeros for each row and making minimal swaps to meet the condition for each row.

Algorithm

Step 1: Create an array trailingZeros to store the number of trailing zeros for each row. Initialize a variable: swaps to count the number of swaps needed.

Step 2: Iterate through each row of the grid. For each row, count the number of trailing zeros by iterating from the last element to the first element. Store the count of trailing zeros in the trailingZeros array.

Step 3: Greedy Row Placement: For each row i (from the first row to the last row): Calculate the required number of trailing zeros for the current row, which is n-1-i.

Step 4: Check if the current row i has the required number of trailing zeros. If not, find a suitable row below it that meets the requirement.

Step 5: If no suitable row is found, return -1 as it's impossible to arrange the grid. If a suitable row is found, swap it upwards to the current position i, and count the number of swaps needed.

Step 6: After all rows are processed, return the total number of swaps counted.

Filename: MinimumSwapsBinaryGrid.java

Output:

 
3

Time Complexity

We iterate through each row (of size n) and for each row, we iterate through its elements (of size n). Therefore, the nested loops result in O(n²) time complexity.

Space Complexity

We use additional space for the trailingZeros array of size n to store the number of trailing zeros for each row. Other than this, we use a few constant extra variables for indices and counters. Therefore, the space complexity is O(n).

Approach 5: Optimized Counting Sort

The optimized counting sort approach leverages counting and sorting techniques to efficiently determine the minimum number of swaps needed to rearrange the rows. By tracking the number of trailing zeros in each row and using counting sort principles, we can determine the correct positions for rows with minimal computational overhead.

Algorithm

Step 1: Create an array trailingZeros of size n to store the count of trailing zeros for each row. Loop through each row i from 0 to n-1. Count the number of consecutive zeros from the end of the row and store this count in trailingZeros[i].

Step 2: Create a counting array to efficiently find rows that can be moved to the current position. Loop through each row i from 0 to n-1. Determine the required number of trailing zeros for the i-th row, which is n - 1 - i.

Step 3: Find the nearest row from the current position that meets this requirement by incrementing rowIndex until a suitable row is found or the end of the grid is reached.

Step 4: Swapping Rows: If a suitable row is found (rowIndex < n), swap rows to move the suitable row up to position i.

Step 5: Use a while loop to perform swaps until rowIndex equals i, and increment the swap count for each swap performed.

Step 6: After processing all rows, return the total number of swaps performed to achieve the desired arrangement.

Filename: MinimumSwapsBinaryGridOCS.java

Output:

 
3

Time Complexity

We iterate through each element of the grid to count trailing zeros, resulting in O(n^2). For each row, in the worst case, we might need to swap it with every other row below it, resulting in O(n²). Therefore, the overall time complexity is O(n²).

Space Complexity

The space complexity for Trailing Zeros Array: O(n), we use an array trailingZeros to store the count of trailing zeros for each row. Hence, overall space complexity is O(n).

Approach 6: Greedy Approach Using Position of Last '1'

The greedy approach involves identifying the position of the last '1' in each row and then rearranging the rows so that each row i satisfies the condition that the last '1' in that row is at position i or earlier.

Algorithm

Step 1: Initialize Position Array: Create an array pos where pos[i] stores the position of the last '1' in row i. If a row has no '1', its position is set to -1.

Step 2: Calculate Positions of Last '1': Iterate through each row of the grid and identify the position of the last '1'. Store these positions in the pos array.

Step 3: Greedy Row Placement: Iterate over each row i from 0 to n-1 and ensure that the row at position i satisfies the condition that the last '1' is at or before position i.

Step 4: If the current row i does not satisfy the condition, find the nearest row below i that does, and swap rows as needed.

Step 5: Perform Swaps: Swap the current row with the suitable row found in the previous step, incrementing the swap count for each swap performed.

Step 6: Check for Valid Configuration: If no suitable row is found, return -1 indicating it is impossible to arrange the grid. Return the total number of swaps performed to achieve the desired arrangement.

Filename: MinimumSwapsBinaryGrid1.java

Output:

 
3

Time Complexity

We iterate through each element of the grid to find the position of the last '1' in each row, this requires n×n=n^2 operations. In the worst case, for each row, we might need to swap it with every other row below it, resulting in O(n) operations for each of the n rows. Therefore, the total complexity is O(n²).

Space Complexity

The space complexity is O(n): The additional space used is mainly for the pos array, which stores the position of the last '1' for each of the n rows.