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This approach involves two distinct passes over the matrix: first to flip each row horizontally by reversing the elements, and second to invert all the elements by replacing 0s with 1s and vice versa.
Time Complexity: O(n^2) as each element is visited once.
Space Complexity: O(1) as no additional space proportional to input size is used.
1#include <vector>
2using namespace std;
3
4vector<vector<int>> flipAndInvertImage(vector<vector<int>>& image) {
5 for (auto& row : image) {
6 reverse(row.begin(), row.end());
7 for (auto& pixel : row) {
8 pixel ^= 1;
9 }
10 }
11 return image;
12}
13
14int main() {
15 vector<vector<int>> image = {{1, 1, 0}, {1, 0, 1}, {0, 0, 0}};
16 image = flipAndInvertImage(image);
17 return 0;
18}In this C++ code, we use STL to reverse each row, and then XOR each element with 1 to invert it.
This approach attempts to combine the flipping and inverting processes in one pass to enhance efficiency, leveraging the symmetry characteristics of the problem.
Time Complexity: O(n^2) given each element may potentially be processed.
Space Complexity: O(1) since modifications are in place.
1public class Solution {
2 public int[][] FlipAndInvertImage(int[][] image) {
int n = image.Length;
foreach (var row in image) {
for (int j = 0; j < (n + 1) / 2; ++j) {
if (row[j] == row[n - 1 - j]) {
row[j] ^= 1;
row[n - 1 - j] ^= 1;
}
}
}
return image;
}
static void Main(string[] args) {
int[][] image = new int[][] { new int[] {1, 1, 0}, new int[]{1, 0, 1}, new int[]{0, 0, 0} };
Solution solution = new Solution();
solution.FlipAndInvertImage(image);
}
}The C# solution uses symmetry efficiently by traversing from both ends of a row, flipping and inverting as needed to minimize operations when the elements are the same.