The algorithm problem - Backtracing pattern in JavaScript
Consider the following backtracing problem: On a 2−dimensional grid, there are 4 types of squares −
1 represents the starting square. There is exactly one starting square.
2 represents the ending square. There is exactly one ending square.
0 represents empty squares we can walk over.
−1 represents obstacles that we cannot walk over.
We are required to write a function that returns the number of 4−directional walks from the starting square to the ending square, that walk over every non−obstacle square exactly once.
Example
const arr = [
[1,0,0,0],
[0,0,0,0],
[0,0,2,-1]
];
const uniquePaths = (arr, count = 0) => {
const dy = [1,−1,0,0], dx = [0,0,1,−1];
const m = arr.length, n = arr[0].length;
const totalZeroes = arr.map(row => row.filter(num => num ===
0).length).reduce((totalZeroes,nextRowZeroes) => totalZeroes +
nextRowZeroes, 0);
const depthFirstSearch = (i, j, covered) => {
if (arr[i][j] === 2){
if (covered === totalZeroes + 1) count++;
return;
};
for (let k = 0; k < 4; k++)
if (i+dy[k] >= 0 && i+dy[k] < m && j+dx[k] >= 0 && j+dx[k] < n
&& arr[i+dy[k]][j+dx[k]] !== −1 ){
arr[i][j] = −1;
depthFirstSearch(i+dy[k],j+dx[k],covered+1);
arr[i][j] = 0;
}
return;
};
for (let row = 0; row < m; row++)
for (let col = 0; col < n; col++)
if (arr[row][col] === 1){
arr[row][col] = −1;
depthFirstSearch(row,col,0);
break;
}
return count;
};
console.log(uniquePaths(arr));Explanation
We set up variables to facilitate four directional iteration when traversing grid, count zeroes in matrix to allow for checking coverage when base condition of recursion reached
Then we set up the DFS (Depth First Search) backtrack function to mark grid with −1 on the active path and to check path length when the finish cell is reached
And lastly, we launch the DFS from the start cell to count all full paths and return the count
Output
And the output in the console will be −
2
Published on 21-Nov-2020 13:56:17
