How do you solve Minimum Operations to Write the Letter Y on a Grid on LeetCode?

Minimum Operations to Write the Letter Y on a Grid (LeetCode #3071) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The grid's structure allows for a clear separation of Y and non-Y cells.

What is the brute force solution for Minimum Operations to Write the Letter Y on a Grid?

The brute force approach for Minimum Operations to Write the Letter Y on a Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible configuration of the grid to see how many operations are needed to convert the cells into the desired format for the letter Y. This is straightforward but inefficient as it doesn't leverage any patterns. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Operations to Write the Letter Y on a Grid?

Minimum Operations to Write the Letter Y on a Grid uses the following concepts: Array, Hash Table, Matrix, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Operations to Write the Letter Y on a Grid?

Always clarify the problem statement and constraints. Discuss your thought process and approach before coding. Consider edge cases, such as the smallest grid size.

What are common mistakes when solving Minimum Operations to Write the Letter Y on a Grid?

Not correctly identifying Y and non-Y cells. Failing to account for the maximum frequency of values.

#3071

Minimum Operations to Write the Letter Y on a Grid

Medium

ArrayHash TableMatrixCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution leverages the structure of the grid and the properties of the letter Y to minimize operations by directly calculating the necessary changes based on counts of values in Y and non-Y cells.

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Algorithm

3 steps

1Step 1: Count occurrences of each value (0, 1, 2) in both Y and non-Y cells.
2Step 2: Determine the maximum frequency of values in Y and non-Y cells.
3Step 3: Calculate the number of operations needed by subtracting the maximum frequency from the total count of Y and non-Y cells.

solution.py22 lines

1# Full working Python code
2
3def min_operations_optimal(grid):
4    n = len(grid)
5    y_count = [0, 0, 0]
6    non_y_count = [0, 0, 0]
7    center = n // 2
8
9    for r in range(n):
10        for c in range(n):
11            if r == c or r + c == n - 1 or c == center:
12                y_count[grid[r][c]] += 1
13            else:
14                non_y_count[grid[r][c]] += 1
15
16    y_operations = len(y_count) - max(y_count)
17    non_y_operations = len(non_y_count) - max(non_y_count)
18
19    return y_operations + non_y_operations
20
21# Example usage
22print(min_operations_optimal([[1,2,2],[1,1,0],[0,1,0]]))  # Output: 3

ℹ

Complexity note: The time complexity is O(n) because we only need to traverse the grid once to count the occurrences of each value in Y and non-Y cells. The space complexity is O(1) since we use a fixed-size array to store counts.

1The grid's structure allows for a clear separation of Y and non-Y cells.
2Counting occurrences efficiently helps minimize operations.

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