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Author: Superklok

JavaScriptFlippingAnImage.md

@@ -1,7 +1,7 @@
-# JavaScript Flipping An Image
-
-## Challenge:
+# JavaScript Flipping an Image
+<br/>
 
+## 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.
@@ -14,56 +14,56 @@ To invert an image means that each `0` is replaced by `1`, and each `1` is repla
 
 * 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:
+### 1<sup>st</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]]`
+```JavaScript
+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]]
+```
 
-### Constraints:
+### 2<sup>nd</sup> Example
+
+```JavaScript
+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]]
+```
 
-`n == image.length`
-<br/>
-`n == image[i].length`
-<br/>
-`1 <= n <= 20`
 <br/>
-`images[i][j]` is either `0` or `1`.
 
-## Solution:
+### Constraints
+
+```JavaScript
+n == image.length
+n == image[i].length
+1 <= n <= 20
+```
+
+- `images[i][j]` is either `0` or `1`.
 
-`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/>
+
+## Solution
+
+```JavaScript
+const flipAndInvertImage = (image) => {
+    for(let row in image) {
+        image[row] = image[row].reverse();
+        image[row] = image[row].map(x=>1-x);
+    }
+
+    return image;
+};
+```
+
 <br/>
 
-## Explanation:
+## Explanation
 
 I've created a function called `flipAndInvertImage` that takes an image as input and returns the modified image. The purpose of this function is to flip the image horizontally and invert the colors of each pixel.
 <br/>

JavaScriptNumberOfIslands.md

@@ -1,121 +1,97 @@
-# JavaScript Number Of Islands
-
-## Challenge:
+# JavaScript Number of Islands
+<br/>
 
+## Challenge
 Given an `m x n` 2D binary grid `grid` which represents a map of `'1'`s (land) and `'0'`s (water), return the number of islands.
 
 An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
-
-### 1<sup>st</sup> Example:
-
-`Input: grid = [`
-<br/>
-`["1","1","1","1","0"],`
-<br/>
-`["1","1","0","1","0"],`
-<br/>
-`["1","1","0","0","0"],`
 <br/>
-`["0","0","0","0","0"]`
 <br/>
-`]`
-<br/>
-`Output: 1`
 
-### 2<sup>nd</sup> Example:
+### 1<sup>st</sup> Example
 
-`Input: grid = [`
-<br/>
-`["1","1","0","0","0"],`
-<br/>
-`["1","1","0","0","0"],`
-<br/>
-`["0","0","1","0","0"],`
-<br/>
-`["0","0","0","1","1"]`
-<br/>
-`]`
-<br/>
-`Output: 3`
+```JavaScript
+Input: grid = [
+['1','1','1','1','0'],
+['1','1','0','1','0'],
+['1','1','0','0','0'],
+['0','0','0','0','0']
+]
+Output: 1
+```
 
-### Constraints:
+### 2<sup>nd</sup> Example
+
+```JavaScript
+Input: grid = [
+['1','1','0','0','0'],
+['1','1','0','0','0'],
+['0','0','1','0','0'],
+['0','0','0','1','1']
+]
+Output: 3
+```
 
-`m == grid.length`
-<br/>
-`n == grid[i].length`
-<br/>
-`1 <= m, n <= 300`
 <br/>
-`grid[i][j]` is `'0'` or `'1'`.
 
-## Solution:
+### Constraints
+
+```JavaScript
+m == grid.length
+n == grid[i].length
+1 <= m, n <= 300
+```
+
+- `grid[i][j]` is `'0'` or `'1'`.
 
-`const numIslands = (grid) => {`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`let count = 0;`
-<br/>
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`for (let i = 0; i < grid.length; i++) {`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`for (let j = 0; j < grid[0].length; j++) {`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`if (grid[i][j] === '1') {`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`count++;`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`dfs(grid, i,j);`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`}`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`}`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`}`
-<br/>
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`return count;`
-<br/>
-`};`
-<br/>
-<br/>
-`const dfs = (grid, row, col) => {`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`if (row < 0 || row >= grid.length || col < 0 || col >= grid[0].length) {`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`return;`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`}`
-<br/>
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`const value = grid[row][col];`
-<br/>
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`if (value === '1') {`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`grid[row][col] = '#';`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`dfs(grid, row + 1, col);`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`dfs(grid, row - 1, col);`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`dfs(grid, row, col + 1);`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`dfs(grid, row, col - 1);`
-<br/>
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;`}`
-<br/>
-`};`
 <br/>
+
+## Solution
+
+```JavaScript
+const numIslands = (grid) => {
+    let count = 0;
+
+    for (let i = 0; i < grid.length; i++) {
+        for (let j = 0; j < grid[0].length; j++) {
+            if (grid[i][j] === '1') {
+                count++;
+                dfs(grid, i,j);
+            }
+        }
+    }
+
+    return count;
+};
+
+const dfs = (grid, row, col) => {
+    if (row < 0 || row >= grid.length || col < 0 || col >= grid[0].length) {
+        return;
+    }
+
+    const value = grid[row][col];
+
+    if (value === '1') {
+        grid[row][col] = '#';
+        dfs(grid, row + 1, col);
+        dfs(grid, row - 1, col);
+        dfs(grid, row, col + 1);
+        dfs(grid, row, col - 1);
+    }
+};
+```
+
 <br/>
 
-## Explanation:
+## Explanation
 
 I've defined two functions: `numIslands` and `dfs`. 
 <br/>
 
-The `numIslands` function takes a grid as input and returns the number of islands present in the grid. It initializes a variable `count` to 0.
+The `numIslands` function takes a grid as input and returns the number of islands present in the grid. It initializes a variable `count` to `0`.
 <br/>
 
-Inside the function, there is a nested loop that iterates over each cell in the grid. If the current cell has a value of '1', indicating it is part of an island, the `count` is incremented and the `dfs` function is called with the current grid and the indices of the current cell.
+Inside the function, there is a nested loop that iterates over each cell in the grid. If the current cell has a value of `'1'`, indicating it is part of an island, the `count` is incremented and the `dfs` function is called with the current grid and the indices of the current cell.
 <br/>
 
 The `dfs` function is a recursive function that performs a depth-first search on the grid. It takes the grid, row index, and column index as input.
@@ -127,7 +103,7 @@ Within the `dfs` function, there is an initial check to see if the current row o
 Next, the value of the current cell is retrieved from the grid.
 <br/>
 
-If the value is '1', indicating an unvisited cell that is part of an island, the value of the cell is changed to '#' to mark it as visited. Then, the `dfs` function is called recursively for the adjacent cells: one cell below, one cell above, one cell to the right, and one cell to the left.
+If the value is `'1'`, indicating an unvisited cell that is part of an island, the value of the cell is changed to `'#'` to mark it as visited. Then, the `dfs` function is called recursively for the adjacent cells: one cell below, one cell above, one cell to the right, and one cell to the left.
 <br/>
 
 This recursive exploration continues until all connected cells of the current island are visited.