# 3071. Minimum Operations to Write the Letter Y on a Grid

## Description

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 cells belonging to the Y are equal.
• All values at cells not belonging to the Y are equal.
• The values at cells belonging to the Y are different from the values at cells not belonging to the Y.

Return the minimum number of operations needed to write the letter Y on the grid given that in one operation you can change the value at any cell to 0, 1, or 2.

Example 1:

Input: grid = [[1,2,2],[1,1,0],[0,1,0]]
Output: 3
Explanation: 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
Explanation: 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.

## Solutions

### 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.

The time complexity is $O(n^2)$, where $n$ is the size of the matrix. The space complexity is $O(1)$.

• class Solution {
public int minimumOperationsToWriteY(int[][] grid) {
int n = grid.length;
int[] cnt1 = new int[3];
int[] cnt2 = new int[3];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
boolean a = i == j && i <= n / 2;
boolean b = i + j == n - 1 && i <= n / 2;
boolean c = j == n / 2 && i >= n / 2;
if (a || b || c) {
++cnt1[grid[i][j]];
} else {
++cnt2[grid[i][j]];
}
}
}
int ans = n * n;
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
if (i != j) {
ans = Math.min(ans, n * n - cnt1[i] - cnt2[j]);
}
}
}
return ans;
}
}

• class Solution {
public:
int minimumOperationsToWriteY(vector<vector<int>>& grid) {
int n = grid.size();
int cnt1[3]{};
int cnt2[3]{};
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
bool a = i == j && i <= n / 2;
bool b = i + j == n - 1 && i <= n / 2;
bool c = j == n / 2 && i >= n / 2;
if (a || b || c) {
++cnt1[grid[i][j]];
} else {
++cnt2[grid[i][j]];
}
}
}
int ans = n * n;
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
if (i != j) {
ans = min(ans, n * n - cnt1[i] - cnt2[j]);
}
}
}
return ans;
}
};

• 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
)


• func minimumOperationsToWriteY(grid [][]int) int {
n := len(grid)
cnt1 := [3]int{}
cnt2 := [3]int{}
for i, row := range grid {
for j, x := range row {
a := i == j && i <= n/2
b := i+j == n-1 && i <= n/2
c := j == n/2 && i >= n/2
if a || b || c {
cnt1[x]++
} else {
cnt2[x]++
}
}
}
ans := n * n
for i := 0; i < 3; i++ {
for j := 0; j < 3; j++ {
if i != j {
ans = min(ans, n*n-cnt1[i]-cnt2[j])
}
}
}
return ans
}

• function minimumOperationsToWriteY(grid: number[][]): number {
const n = grid.length;
const cnt1: number[] = Array(3).fill(0);
const cnt2: number[] = Array(3).fill(0);
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
const a = i === j && i <= n >> 1;
const b = i + j === n - 1 && i <= n >> 1;
const c = j === n >> 1 && i >= n >> 1;
if (a || b || c) {
++cnt1[grid[i][j]];
} else {
++cnt2[grid[i][j]];
}
}
}
let ans = n * n;
for (let i = 0; i < 3; ++i) {
for (let j = 0; j < 3; ++j) {
if (i !== j) {
ans = Math.min(ans, n * n - cnt1[i] - cnt2[j]);
}
}
}
return ans;
}