What is the time complexity of Minimum Operations to Write the Letter Y on a Grid?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Minimum Operations to Write the Letter Y on a Grid?

This problem uses Array, Hash Table, Matrix, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Operations to Write the Letter Y on a Grid?

Insight 1: The problem's core is about finding optimal values for the Y and non-Y cells among the available values (0, 1, and 2) to minimize changes. Insight 2: The grid's size is relatively small (up to 49x49), so a brute-force approach by trying all value combinations for Y and non-Y cells is feasible.

Which companies ask LeetCode 3071. Minimum Operations to Write the Letter Y on a Grid in interviews?

This problem is frequently asked at Capital One, Visa, Zeta, ZipRecruiter.

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Minimum Operations to Write the Letter Y on a Grid - LeetCode 3071 Solution

Minimum Operations to Write the Letter Y on a Grid - Complete Solution Guide

Minimum Operations to Write the Letter Y on a Grid is LeetCode problem 3071, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 0-indexed n x n grid where n is odd, and grid[r][c] is 0 , 1 , or 2 . We say that a cell belongs to the Letter Y if it belongs to one of the following: The diagonal starting at the top-left cell and ending at the center cell of the grid. The diagonal starting at the top-right cell and ending at the center cell of the grid. The vertical line starting at the center cell and ending at the bottom border of the grid. The Letter Y is written on the grid if and only if: All values at ce

Detailed Explanation

The problem asks us to find the minimum number of operations to transform a given n x n grid into a 'Letter Y' pattern. The grid contains only the values 0, 1, or 2. A cell belongs to the Letter Y if it is part of either of the diagonals from top-left or top-right to the center, or the vertical line from the center to the bottom. The 'Letter Y' pattern is achieved when all cells belonging to the Y have the same value, all cells not belonging to the Y have the same value, and these two values are different. We need to determine the fewest changes needed in the grid to satisfy these conditions.

Solution Approach

The provided solution employs a brute-force approach combined with counting. It first identifies which cells belong to the Letter Y and which don't. Then, it counts the number of times each value (0, 1, 2) appears in both the Y cells and the non-Y cells. Finally, it iterates through all possible combinations of Y and non-Y values, calculates the number of operations needed for each combination, and returns the minimum number of operations found.

Step-by-Step Algorithm

Step 1: Determine the center of the grid: `center = n // 2`.
Step 2: Initialize two arrays, `y_counts` and `non_y_counts`, of size 3 to store the counts of each value (0, 1, 2) in the Y and non-Y cells, respectively.
Step 3: Iterate through each cell of the grid (r, c).
Step 4: Determine if the current cell (r, c) belongs to the Letter Y. This involves checking if it lies on the diagonal from top-left, the diagonal from top-right, or the vertical line from the center. The condition is: `(r == c and r <= center) or (r + c == n - 1 and r <= center) or (c == center and r >= center)`.
Step 5: Increment the corresponding count in `y_counts` or `non_y_counts` based on whether the cell belongs to the Y and its value (`grid[r][c]`).
Step 6: Calculate the total number of Y cells and non-Y cells.
Step 7: Initialize `min_ops` to a large value (e.g., n * n).
Step 8: Iterate through all possible combinations of Y value (`y_val`) and non-Y value (`non_y_val`) from 0 to 2. Ensure `y_val` and `non_y_val` are different.
Step 9: Calculate the number of operations needed to make all Y cells equal to `y_val`: `ops_y = total_y_cells - y_counts[y_val]`.
Step 10: Calculate the number of operations needed to make all non-Y cells equal to `non_y_val`: `ops_non_y = total_non_y_cells - non_y_counts[non_y_val]`.
Step 11: Calculate the total number of operations for this combination: `current_ops = ops_y + ops_non_y`.
Step 12: Update `min_ops` if `current_ops` is smaller: `min_ops = min(min_ops, current_ops)`.
Step 13: Return `min_ops`.

Key Insights

Insight 1: The problem's core is about finding optimal values for the Y and non-Y cells among the available values (0, 1, and 2) to minimize changes.
Insight 2: The grid's size is relatively small (up to 49x49), so a brute-force approach by trying all value combinations for Y and non-Y cells is feasible.
Insight 3: Counting the occurrences of each value (0, 1, and 2) within the Y and non-Y regions helps to efficiently calculate the number of operations needed for each combination.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, Matrix, Counting.

Companies

Asked at: Capital One, Visa, Zeta, ZipRecruiter.

Minimum Operations to Write the Letter Y on a Grid | LeetCode 3071

Medium

Array Hash Table Matrix Counting

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minimumOperationsToWriteY(int[][] grid) {
        int n = grid.length;
        int center = n / 2;

        int[] y_counts = new int[3];
        int[] non_y_counts = new int[3];

        for (int r = 0; r < n; r++) {
            for (int c = 0; c < n; c++) {
                boolean is_y_cell = (r == c && r <= center) ||
                            (r + c == n - 1 && r <= center) ||
                            (c == center && r >= center);
                
                int val = grid[r][c];
                if (is_y_cell) {
                    y_counts[val] += 1;
                } else {
                    non_y_counts[val] += 1;
                }
            }
        }

        int total_y_cells = 0;
        for (int count : y_counts) {
            total_y_cells += count;
        }
        int total_non_y_cells = n * n - total_y_cells;
        
        int min_ops = n * n;

        for (int y_val = 0; y_val < 3; y_val++) {
            for (int non_y_val = 0; non_y_val < 3; non_y_val++) {
                if (y_val == non_y_val) {
                    continue;
                }
                
                int ops_y = total_y_cells - y_counts[y_val];
                int ops_non_y = total_non_y_cells - non_y_counts[non_y_val];
                
                int current_ops = ops_y + ops_non_y;
                min_ops = Math.min(min_ops, current_ops);
            }
        }
        
        return min_ops;
    }
}

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Problem Info

Description

Problem 3071: Minimum Operations to Write the Letter Y on a Grid - Detailed explanation and solution approaches.

Topics

Array Hash Table Matrix Counting

Companies

Capital OneVisaZetaZipRecruiter

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:3071

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

13 steps

1
Step 1: Determine the center of the grid: `center = n // 2`.
2
Step 2: Initialize two arrays, `y_counts` and `non_y_counts`, of size 3 to store the counts of each value (0, 1, 2) in the Y and non-Y cells, respectively.
3
Step 3: Iterate through each cell of the grid (r, c).
4
Step 4: Determine if the current cell (r, c) belongs to the Letter Y. This involves checking if it lies on the diagonal from top-left, the diagonal from top-right, or the vertical line from the center. The condition is: `(r == c and r <= center) or (r + c == n - 1 and r <= center) or (c == center and r >= center)`.
5
Step 5: Increment the corresponding count in `y_counts` or `non_y_counts` based on whether the cell belongs to the Y and its value (`grid[r][c]`).
6
Step 6: Calculate the total number of Y cells and non-Y cells.
7
Step 7: Initialize `min_ops` to a large value (e.g., n * n).
8
Step 8: Iterate through all possible combinations of Y value (`y_val`) and non-Y value (`non_y_val`) from 0 to 2. Ensure `y_val` and `non_y_val` are different.
9
Step 9: Calculate the number of operations needed to make all Y cells equal to `y_val`: `ops_y = total_y_cells - y_counts[y_val]`.
10
Step 10: Calculate the number of operations needed to make all non-Y cells equal to `non_y_val`: `ops_non_y = total_non_y_cells - non_y_counts[non_y_val]`.
11
Step 11: Calculate the total number of operations for this combination: `current_ops = ops_y + ops_non_y`.
12
Step 12: Update `min_ops` if `current_ops` is smaller: `min_ops = min(min_ops, current_ops)`.
13
Step 13: Return `min_ops`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is primarily determined by the nested loops used to iterate through the grid. The rest of the operations, such as calculating counts and finding the minimum operations, take constant time. The space complexity is constant because the additional space used doesn't depend on the input size 'n'.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

Although less efficient for this specific problem due to the small constraint size, a dynamic programming approach could be considered for a generalized version of the problem with different shapes or patterns. It would likely involve storing the minimum operations to form a partial pattern up to a certain row or column.

Time: Potentially higher than O(n^2) depending on how it's implemented. Likely O(n^3) or higher.Space: O(n^2) to store the DP table.

Trade-offs: DP would be an overkill for this particular problem. It could become useful if the constraints were significantly larger or the target shape was more complex, requiring more complex state management.

✨

Key Insights

1Insight 1: The problem's core is about finding optimal values for the Y and non-Y cells among the available values (0, 1, and 2) to minimize changes.
2Insight 2: The grid's size is relatively small (up to 49x49), so a brute-force approach by trying all value combinations for Y and non-Y cells is feasible.
3Insight 3: Counting the occurrences of each value (0, 1, and 2) within the Y and non-Y regions helps to efficiently calculate the number of operations needed for each combination.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: Grids where all cells already form a valid Y pattern. The algorithm will still correctly calculate the minimum operations needed, which would be 0 in this case.
Edge case 2: Small grids (e.g., 3x3) with very few non-Y cells. The algorithm handles this correctly because it accurately counts the occurrences of each value in both regions and chooses the optimal combination.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly identifying the cells that belong to the Letter Y. A careful examination of the indices and the constraints (r == c, r + c == n-1, c == center) is crucial.

✗Common mistake 2: Forgetting to ensure that the Y value and the non-Y value are different. The problem specifies this as a requirement, and failing to enforce it will lead to incorrect results.

✗Common mistake 3: Not considering all possible combinations of values for the Y and non-Y sections (0, 1, and 2). Missing a combination could lead to a suboptimal solution.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Image processing and pattern recognition. This problem demonstrates a simplified version of identifying and correcting patterns in images. For example, identifying and correcting defects in manufacturing processes using image analysis.

🏦Real-world scenario 2: Data cleaning and standardization. In data analysis, similar logic can be used to identify and correct inconsistencies in datasets, where certain data points should adhere to a specific pattern or rule.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 12, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Operations to Write the Letter Y on a Grid"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?