## Saturday, May 26, 2018

### LeetCode - 832 Flipping an Image

https://leetcode.com/problems/flipping-an-image/description/
Given a binary matrix `A`, we want to flip the image horizontally, then invert it, and return the resulting image.
To flip an image horizontally means that each row of the image is reversed.  For example, flipping `[1, 1, 0]` horizontally results in `[0, 1, 1]`.
To invert an image means that each `0` is replaced by `1`, and each `1` is replaced by `0`. For example, inverting `[0, 1, 1]` results in `[1, 0, 0]`.
Example 1:
```Input: [[1,1,0],[1,0,1],[0,0,0]]
Output: [[1,0,0],[0,1,0],[1,1,1]]
Explanation: First reverse each row: [[0,1,1],[1,0,1],[0,0,0]].
Then, invert the image: [[1,0,0],[0,1,0],[1,1,1]]
```
Example 2:
```Input: [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]]
Output: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]
Explanation: First reverse each row: [[0,0,1,1],[1,0,0,1],[1,1,1,0],[0,1,0,1]].
Then invert the image: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]
```
Notes:
• `1 <= A.length = A[0].length <= 20`
• `0 <= A[i][j] <= 1`
We can do this in place. In each row, the `i`th value from the left is equal to the inverse of the `i`th value from the right.
We use `(C+1) / 2` (with floor division) to iterate over all indexes `i` in the first half of the row, including the center.
In Python, the shortcut `row[~i] = row[-i-1] = row[len(row) - 1 - i]` helps us find the `i`th value of the row, counting from the right.

public int[][] flipAndInvertImage(int[][] A) {
int C = A[0].length;
for (int[] row: A)
for (int i = 0; i < (C + 1) / 2; ++i) {
int tmp = row[i] ^ 1;
row[i] = row[C - 1 - i] ^ 1;
row[C - 1 - i] = tmp;
}

return A;
}