Add JavaScriptFlippingAnImage.md

Author: Superklok

JavaScriptFlippingAnImage.md

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+# JavaScript Flipping An Image
+
+## Challenge:
+
+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.
+<br/>
+
+* 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`.
+<br/>
+
+* For example, inverting `[0,1,1]` results in `[1,0,0]`.
+
+### 1<sup>st</sup> Example:
+
+`Input: image = [[1,1,0],[1,0,1],[0,0,0]]`
+<br/>
+`Output: [[1,0,0],[0,1,0],[1,1,1]]`
+<br/>
+`Explanation: First reverse each row: [[0,1,1],[1,0,1],[0,0,0]].`
+<br/>
+`Then, invert the image: [[1,0,0],[0,1,0],[1,1,1]]`
+
+### 2<sup>nd</sup> Example:
+
+`Input: image = [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]]`
+<br/>
+`Output: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]`
+<br/>
+`Explanation: First reverse each row: [[0,0,1,1],[1,0,0,1],[1,1,1,0],[0,1,0,1]].`
+<br/>
+`Then invert the image: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]`
+
+### Constraints:
+
+`n == image.length`
+<br/>
+`n == image[i].length`
+<br/>
+`1 <= n <= 20`
+<br/>
+`images[i][j]` is either `0` or `1`.
+
+## Solution:
+
+`const flipAndInvertImage = (image) => {`
+<br/>
+&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`for(let row in image) {`
+<br/>
+&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`image[row] = image[row].reverse();`
+<br/>
+&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`image[row] = image[row].map(x=>1-x);`
+<br/>
+&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`}`
+<br/>
+<br/>
+&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`return image;`
+<br/>
+`};`
+<br/>
+<br/>