How do you solve Minimum Swaps to Arrange a Binary Grid on LeetCode?

Minimum Swaps to Arrange a Binary Grid (LeetCode #1536) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: The rightmost '1' in each row determines the validity of the grid.

What is the time complexity of Minimum Swaps to Arrange a Binary Grid?

The optimal solution for Minimum Swaps to Arrange a Binary Grid runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to sorting the maxRight array, and the space complexity is O(n) for storing the maxRight values.

What is the brute force solution for Minimum Swaps to Arrange a Binary Grid?

The brute force approach for Minimum Swaps to Arrange a Binary Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking all possible row swaps to see if we can achieve a valid grid. This method is straightforward but inefficient, as it may require many swaps to find a valid configuration. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Swaps to Arrange a Binary Grid?

Minimum Swaps to Arrange a Binary Grid uses the following concepts: Array, Greedy, Matrix. The recommended patterns to study are: Greedy, Sorting, Array.

What are the interview tips for Minimum Swaps to Arrange a Binary Grid?

Clearly explain your thought process while coding. Consider edge cases, such as grids that are already valid. Practice explaining your code to someone else to reinforce your understanding.

What are common mistakes when solving Minimum Swaps to Arrange a Binary Grid?

Not checking if the grid can be valid before counting swaps. Confusing the indices while counting swaps.

#1536

Minimum Swaps to Arrange a Binary Grid

Medium

ArrayGreedyMatrixGreedySortingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal approach leverages the sorted order of the rightmost '1' indices to minimize swaps. By ensuring that each row's maxRight value is less than or equal to its index, we can efficiently count the necessary swaps.

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Algorithm

4 steps

1Step 1: Calculate the rightmost '1' for each row and store it in an array maxRight.
2Step 2: Sort the maxRight array.
3Step 3: For each index i in the sorted maxRight, ensure that maxRight[i] <= i. If not, return -1.
4Step 4: Count the number of swaps needed to arrange the rows based on the maxRight values.

solution.py18 lines

1def minSwaps(grid):
2    n = len(grid)
3    maxRight = [0] * n
4    for i in range(n):
5        for j in range(n - 1, -1, -1):
6            if grid[i][j] == 1:
7                maxRight[i] = j
8                break
9    sortedMaxRight = sorted(maxRight)
10    for i in range(n):
11        if sortedMaxRight[i] > i:
12            return -1
13    swaps = 0
14    for i in range(n):
15        while maxRight[i] != sortedMaxRight[i]:
16            swaps += 1
17            maxRight[i], maxRight[i + 1] = maxRight[i + 1], maxRight[i]
18    return swaps

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the maxRight array, and the space complexity is O(n) for storing the maxRight values.

1The rightmost '1' in each row determines the validity of the grid.
2Sorting the maxRight values helps in minimizing the number of swaps.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.