Flipping an Image | LeetCode 832 | C

2023-09-25
#LeetCode #C #Array #Two Pointers #Matrix #Simulation
LeetCode

Given an n x n binary matrix image, 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:

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Input: image = [[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:

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Input: image = [[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]]

Constraints:

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n == image.length
n == image[i].length
1 <= n <= 20
images[i][j] is either 0 or 1.

Approach

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Flip the image by the method of two pointers.

Algorithm

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Go through the array and compare two elements. If the elements are equal, invert the two elements. If the two elements are not equal, do nothing.

Implementation

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int** flipAndInvertImage(int** image, int imageSize, int* imageColSize, int* returnSize, int** returnColumnSizes){
    int **ret = malloc(imageSize * sizeof(int *));
    for(int i = 0; i< imageSize;i++){
        int left = 0;
        int right = imageSize - 1; 
        while(left<=right){
            if(left == right){
                image[i][left] = !image[i][left];
                break;
            }else if(image[i][left] == image[i][right]){
                image[i][left] = !image[i][left];
                image[i][right] = !image[i][right];
            }
            left++;
            right--;
        }
    }
    *returnSize = imageSize;
	*returnColumnSizes = imageColSize;
    return image;
}