#3071

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Minimum Operations to Write the Letter Y on a Grid

Find the minimum number of operations to write the letter Y on a grid by altering cell values.

Problem Statement

You are given an odd-sized n x n grid where each cell contains one of the values 0, 1, or 2. Your task is to determine the minimum number of operations required to convert the grid into the shape of the letter Y. The letter Y must consist of cells with the same value, and the grid must be altered with the fewest changes possible.

The shape of the letter Y in the grid is defined by specific cells that form a Y. Your goal is to modify the grid with the minimum operations to make these cells uniform in value while leaving the other cells unchanged. The number of operations is the sum of the changes made to the grid cells to create this Y shape.

Examples

Example 1

Input: grid = [[1,2,2],[1,1,0],[0,1,0]]

Output: 3

We can write Y on the grid by applying the changes highlighted in blue in the image above. After the operations, all cells that belong to Y, denoted in bold, have the same value of 1 while those that do not belong to Y are equal to 0. It can be shown that 3 is the minimum number of operations needed to write Y on the grid.

Example 2

Input: grid = [[0,1,0,1,0],[2,1,0,1,2],[2,2,2,0,1],[2,2,2,2,2],[2,1,2,2,2]]

Output: 12

We can write Y on the grid by applying the changes highlighted in blue in the image above. After the operations, all cells that belong to Y, denoted in bold, have the same value of 0 while those that do not belong to Y are equal to 2. It can be shown that 12 is the minimum number of operations needed to write Y on the grid.

Constraints

3 <= n <= 49
n == grid.length == grid[i].length
0 <= grid[i][j] <= 2
n is odd.

Solution Approach

Identify the Y Shape

The first step is to identify the cells in the grid that will make up the Y shape. This is done by analyzing the structure of the Y and marking the grid cells that belong to it.

Efficient Value Matching

Once the Y shape cells are identified, the next task is to determine the minimum number of changes needed to make the value of those cells uniform. Using hash table lookups, we can count the occurrences of each value (0, 1, 2) in the Y shape, which allows us to choose the most frequent value for uniformity and minimize changes.

Calculate Minimum Operations

The final step is to compute the number of operations required. This is done by calculating how many cells need to be changed to match the most frequent value within the Y shape, ensuring the least number of operations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the implementation of the Y shape identification and hash table lookup. The grid size is n x n, where n is odd, so the time complexity is O(n^2) due to scanning the grid. Space complexity is O(n^2) for storing the grid and hash table data.

What Interviewers Usually Probe

Look for the candidate's ability to identify the Y-shaped region efficiently.
Assess how well the candidate uses hash tables for counting occurrences and minimizing operations.
Evaluate the candidate's approach to calculating the number of operations needed and how they minimize changes.

Common Pitfalls or Variants

Common pitfalls

Overlooking the Y shape's exact structure, leading to incorrect cell selection.
Choosing a value that isn't the most frequent within the Y shape, resulting in unnecessary changes.
Misunderstanding the problem's constraints, such as the odd grid size or incorrect range of values.

Follow-up variants

The grid size could vary, but the problem will always involve scanning the grid and identifying the Y shape.
The problem could be extended to allow for different target shapes, requiring an adaptation of the Y-shape identification process.
The grid values could be extended beyond 0, 1, and 2, requiring an updated method for value counting and selection.

FAQ

What is the minimum number of operations to write the letter Y on a grid?

The minimum number of operations is the number of changes required to make all cells that belong to the Y shape uniform in value.

How do I identify the Y shape on the grid?

The Y shape consists of cells arranged in a specific pattern, usually forming a central vertical line and two diagonal arms that meet at the top. You need to scan the grid to locate these positions.

How does hash table lookup help solve the problem?

By counting the occurrences of each value within the Y shape using a hash table, we can determine which value is most frequent, minimizing the number of changes needed to make all Y cells uniform.

What if the grid contains more than three values?

If the grid contains additional values, you'll need to adjust the hash table logic to accommodate them, but the approach of counting and selecting the most frequent value remains the same.

Can the grid size change in the problem?

Yes, the grid size can vary, but the basic approach of scanning the grid and using hash tables for value selection remains effective for different sizes.

terminal

Solution

Solution 1: Counting

We use two arrays of length 3, `cnt1` and `cnt2`, to record the counts of cell values that belong to `Y` and do not belong to `Y`, respectively. Then we enumerate `i` and `j`, which represent the values of cells that belong to `Y` and do not belong to `Y`, respectively, to calculate the minimum number of operations.

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class Solution:
    def minimumOperationsToWriteY(self, grid: List[List[int]]) -> int:
        n = len(grid)
        cnt1 = Counter()
        cnt2 = Counter()
        for i, row in enumerate(grid):
            for j, x in enumerate(row):
                a = i == j and i <= n // 2
                b = i + j == n - 1 and i <= n // 2
                c = j == n // 2 and i >= n // 2
                if a or b or c:
                    cnt1[x] += 1
                else:
                    cnt2[x] += 1
        return min(
            n * n - cnt1[i] - cnt2[j] for i in range(3) for j in range(3) if i != j
        )

Continue Practicing

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