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

The problem asks us to determine the minimum number of operations to convert a given n x n grid into a representation of the letter Y, under specific conditions. The grid contains integers 0, 1, or 2.

leetcodemediumarrayhash-tablematrixcounting

LeetCode Problem 3071

Difficulty: 🟡 Medium
Topics: Array, Hash Table, Matrix, Counting

Solution

Problem Understanding

The problem asks us to determine the minimum number of operations to convert a given n x n grid into a representation of the letter Y, under specific conditions. The grid contains integers 0, 1, or 2. The cells that belong to the Y consist of two diagonals starting from the top corners to the center, and a vertical line extending from the center to the bottom of the grid. The cells not in the Y are every other cell. To successfully "write" the letter Y:

  1. All cells that belong to the Y must have the same value.
  2. All cells that do not belong to the Y must have the same value.
  3. The Y cells and non-Y cells must have different values.

An operation is changing any cell to 0, 1, or 2. We need to minimize the total number of operations to satisfy the Y constraints.

The constraints tell us that the grid size is small to medium (3 <= n <= 49) and always odd, which simplifies finding the exact center for constructing the Y. All grid values are integers from 0 to 2. Edge cases include grids that are already uniform, grids where all cells are the same, and minimal size grids (3x3), which can test the correctness of center and diagonal calculations.

Approaches

Brute Force

A naive approach would be to iterate over all possible assignments for Y and non-Y cells, trying every combination of values from 0, 1, 2. For each assignment, count how many changes are needed and keep the minimum. Since there are 3 choices for the Y and 2 remaining choices for the non-Y cells (they must differ), this gives 6 combinations. For each combination, we scan the grid to count changes, giving O(n^2) per combination. This brute-force method is feasible for our constraints because n is at most 49, but it is not elegant.

Optimal Insight

The key observation is that there are only 3 possible values (0, 1, 2) and two groups of cells (Y and non-Y). We can precompute the frequency of each value in both groups, then calculate the minimum operations for each possible assignment:

  • Let Y_count[v] be the number of Y cells with value v.
  • Let nonY_count[v] be the number of non-Y cells with value v.
  • For a candidate assignment (Y_val, nonY_val), the required operations are (Y_cells - Y_count[Y_val]) + (nonY_cells - nonY_count[nonY_val]).
  • Try all valid (Y_val, nonY_val) pairs with Y_val != nonY_val and take the minimum.

This avoids redundant recalculation and reduces the problem to simple counting.

Approach Time Complexity Space Complexity Notes
Brute Force O(6 * n^2) = O(n^2) O(1) Try all value pairs, scan grid each time
Optimal O(n^2) O(1) Precompute counts, evaluate all valid Y/non-Y value pairs

Algorithm Walkthrough

  1. Identify all the cells belonging to the letter Y. Iterate over the grid: a cell (r, c) is in Y if it is on the top-left to center diagonal, top-right to center diagonal, or the vertical line from the center to the bottom.
  2. Count the occurrences of each value (0, 1, 2) in the Y cells and non-Y cells separately.
  3. Calculate the total number of Y and non-Y cells.
  4. Enumerate all valid assignments of values (Y_val, nonY_val) with Y_val != nonY_val.
  5. For each assignment, compute the number of operations as (Y_total - Y_count[Y_val]) + (nonY_total - nonY_count[nonY_val]).
  6. Return the minimum number of operations across all assignments.

Why it works: The algorithm guarantees correctness because it exhaustively considers all valid assignments of values to the two groups. Counting ensures we only change cells that do not match the assigned value, which minimizes the number of operations.

Python Solution

from typing import List

class Solution:
    def minimumOperationsToWriteY(self, grid: List[List[int]]) -> int:
        n = len(grid)
        center = n // 2
        
        Y_cells = []
        nonY_cells = []
        
        for r in range(n):
            for c in range(n):
                if (r <= center and (r == c or r + c == n - 1)) or (r >= center and c == center):
                    Y_cells.append(grid[r][c])
                else:
                    nonY_cells.append(grid[r][c])
        
        Y_count = [0, 0, 0]
        nonY_count = [0, 0, 0]
        
        for v in Y_cells:
            Y_count[v] += 1
        for v in nonY_cells:
            nonY_count[v] += 1
        
        Y_total = len(Y_cells)
        nonY_total = len(nonY_cells)
        
        min_ops = float('inf')
        
        for Y_val in range(3):
            for nonY_val in range(3):
                if Y_val == nonY_val:
                    continue
                ops = (Y_total - Y_count[Y_val]) + (nonY_total - nonY_count[nonY_val])
                min_ops = min(min_ops, ops)
        
        return min_ops

Explanation: First, we separate Y and non-Y cells. Then, we count how many of each value exist in both groups. Using these counts, we can quickly compute the number of operations required for every possible valid assignment of values to the Y and non-Y cells, ensuring that we get the minimum possible.

Go Solution

func minimumOperationsToWriteY(grid [][]int) int {
    n := len(grid)
    center := n / 2

    Y_count := [3]int{}
    nonY_count := [3]int{}

    Y_total := 0
    nonY_total := 0

    for r := 0; r < n; r++ {
        for c := 0; c < n; c++ {
            inY := (r <= center && (r == c || r+c == n-1)) || (r >= center && c == center)
            if inY {
                Y_count[grid[r][c]]++
                Y_total++
            } else {
                nonY_count[grid[r][c]]++
                nonY_total++
            }
        }
    }

    minOps := n * n
    for Y_val := 0; Y_val < 3; Y_val++ {
        for nonY_val := 0; nonY_val < 3; nonY_val++ {
            if Y_val == nonY_val {
                continue
            }
            ops := (Y_total - Y_count[Y_val]) + (nonY_total - nonY_count[nonY_val])
            if ops < minOps {
                minOps = ops
            }
        }
    }

    return minOps
}

Explanation: The Go solution mirrors the Python logic. Arrays are used instead of lists for counting, and we use a nested loop to evaluate all valid value assignments. Go-specific considerations include predefining the array size and using standard loops instead of Python list comprehensions.

Worked Examples

Example 1:

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

Y cells: positions (0,0),(0,2),(1,1),(2,1) with values [1,2,1,1]

Non-Y cells: positions (0,1),(1,0),(1,2),(2,0),(2,2) with values [2,1,0,0,0]

Counts:

Y_count = [0:0, 1:3, 2:1]

nonY_count = [0:3,1:1,2:1]

Try all valid (Y_val, nonY_val) pairs:

  • (1,0) → ops = (4-3) + (5-3) = 1+2=3
  • (1,2) → ops = 1+4=5
  • (2,0) → ops = 3+2=5
  • ... → minimum = 3

Output = 3

Example 2:

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]]

Following the same process yields minimum operations = 12.

Complexity Analysis

Measure Complexity Explanation
Time O(n^2) We scan the grid once to classify cells and count values, and then try at most 6 assignments.
Space O(1) We only store counts for values 0,1,2 in two arrays and temporary lists of cell values.

The quadratic complexity is acceptable because n ≤ 49, giving at most ~2400 cells.

Test Cases

# Provided examples
assert Solution().minimumOperationsToWriteY([[1,2,2],[1,1,0],[0,1,0]]) == 3  # Example 1