- Difficulty: Medium
- Tags: LeetCode, Medium, Array, Hash Table, Counting, Matrix, leetcode-3071, O(n^2), O(1)
Problem
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.