Bit Manipulation - Flip and Invert Image

Given a binary matrix image, flip each row horizontally and then invert (0→1, 1→0). Return the resulting image.

Input: [[1,1,0],[1,0,1],[0,0,0]] → Output: [[1,0,0],[0,1,0],[1,1,1]]

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]]

For each row, use two pointers from both ends. Swap elements while XOR-ing with 1 to flip and invert simultaneously. When pointers meet at the middle (odd length), just flip the middle element.

Use two pointers lo and hi for each row.
While lo < hi: swap arr[lo] and arr[hi] and XOR both with 1.
When lo == hi (middle of odd-length row): flip arr[lo] ^= 1.
Process all rows.

class Solution { public int[][] flipAndInvertImage(int[][] image) { for (int[] row : image) { int lo = 0, hi = row.length - 1; while (lo <= hi) { int tmp = row[lo] ^ 1; row[lo] = row[hi] ^ 1; row[hi] = tmp; lo++; hi--; } } return image; } }

Time Complexity: O(M*N)
Space Complexity: O(1) in-place