832.flipping-an-image.cpp

/*
 * @lc app=leetcode id=832 lang=cpp
 *
 * [832] Flipping an Image
 *
 * https://leetcode.com/problems/flipping-an-image/description/
 *
 * algorithms
 * Easy (69.75%)
 * Total Accepted:    57.6K
 * Total Submissions: 82.3K
 * Testcase Example:  '[[1,1,0],[1,0,1],[0,0,0]]'
 *
 * 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
 * 
 * 
 */
class Solution {
public:
    vector<vector<int>> flipAndInvertImage(vector<vector<int>>& A) {
        int R = A.size();
        int C = A[0].size();
        for(int i=0;i<R;i++){
            for(int j=0;j<=C/2;j++){
                if(j > C-1-j)
                    continue;
                else if(j == C-1-j)
                    A[i][j] = A[i][j]^1;
                else if(A[i][j] == A[i][C-1-j]){
                    A[i][j] = A[i][j]^1;
                    A[i][C-1-j] = A[i][C-1-j]^1;
                }
            }
        }
        return A;
    }
};