LeetCode-in-Java

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

Medium

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.

Solution

public class Solution {
    public int minimumOperationsToWriteY(int[][] arr) {
        int n = arr.length;
        int[] cnt1 = new int[3];
        int[] cnt2 = new int[3];
        int x = n / 2;
        int y = n / 2;
        for (int j = x; j < n; j++) {
            cnt1[arr[j][y]]++;
            arr[j][y] = 3;
        }
        for (int j = x; j >= 0; j--) {
            if (arr[j][j] != 3) {
                cnt1[arr[j][j]]++;
            }
            arr[j][j] = 3;
        }
        for (int j = x; j >= 0; j--) {
            if (arr[j][n - 1 - j] != 3) {
                cnt1[arr[j][n - 1 - j]]++;
            }
            arr[j][n - 1 - j] = 3;
        }
        for (int[] ints : arr) {
            for (int j = 0; j < n; j++) {
                if (ints[j] != 3) {
                    cnt2[ints[j]]++;
                }
            }
        }
        int s1 = cnt1[0] + cnt1[1] + cnt1[2];
        int s2 = cnt2[0] + cnt2[1] + cnt2[2];
        int min = Integer.MAX_VALUE;
        for (int i = 0; i <= 2; i++) {
            for (int j = 0; j <= 2; j++) {
                if (i != j) {
                    min = Math.min(s1 - cnt1[i] + s2 - cnt2[j], min);
                }
            }
        }
        return min;
    }
}

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