# Solution For Number Of Good Paths

Problem Statement:

Given a positive integer n, you need to count all possible distinct good paths in a n x n grid. A good path is a path from the upper-left cell (1, 1) to the lower-right cell (n, n) that satisfies the following conditions:

The path only contains cells in the diagonal direction (i.e. cells (i, j) such that i==j).
For every pair of consecutive cells (i, j) and (i+1, j+1) in the path, either the color of cell (i, j) and cell (i+1, j+1) must be the same, or cell(i, j) and cell(i+1, j+1) must be both white (i.e. have a value of 0).
Two paths are considered distinct if and only if there is at least one cell that belongs to one path and does not belong to the other. For example, the paths (1, 1) -> (2, 2) -> (3, 3) and (1, 1) -> (3, 3) are considered distinct.
Return the number of distinct good paths modulo 109+7.

Example:

Input: n = 2
Output: 1
Explanation: A good path can only be diagonal cells (1, 1) -> (2, 2) and (2, 2) -> (1, 1).

Solution:

First of all, we need to understand what a good path is. Given the description, a good path is a path that starts from (1,1) and ends at (n,n) and only contains diagonal moves and satisfies the following two conditions:

For every pair of consecutive cells (i,j) and (i+1,j+1) in the path, either the color of cell (i,j) and cell (i+1,j+1) must be the same, or cell(i,j) and cell(i+1,j+1) must be both white (i.e. have a value of 0).
The path only contains cells in the diagonal direction (i.e. cells (i,j) such that i==j).

To count the number of good paths, we can use dynamic programming. Let’s define dp[i][j][k] as the number of good paths that start from cell (i,j) and ends at cell (n,n) and the color of cell (i,j) is k. Here, 0 represents a white cell, and 1 represents a black cell. We can initialize dp[n][n][0] and dp[n][n][1] as 1 because there is only one possible path that starts and ends at cell (n,n).

To compute dp[i][j][k], we need to consider two cases:

If cell (i,j) has the same color as cell (i+1,j+1), then we can only move diagonally to cell (i+1,j+1) and stay in the same color.
dp[i][j][k] = dp[i+1][j+1][k]

If cell (i,j) and cell (i+1,j+1) are both white, then we can either move diagonally to cell (i+1,j+1) and stay in the same color or move diagonally to cell (i+1,j+1) and switch colors.
dp[i][j][k] = dp[i+1][j+1][k] + dp[i+1][j+1][1-k]

The answer will be dp[1][1][0] + dp[1][1][1] (since we can start with either white or black cell).

The time complexity of the above solution is O(n^3), and the space complexity is O(n^3) as well. However, we can optimize the space complexity to O(n^2) by keeping only the previous and current rows of the dp array.

## Step by Step Implementation For Number Of Good Paths

```There are a few different ways to approach this problem. One way would be to use a graph traversal algorithm, such as DFS or BFS, to find all of the possible paths through the grid. Once all of the paths have been found, you can then count how many of them are "good" paths (i.e. paths that go through all of the required nodes).

Another approach would be to use a dynamic programming algorithm to solve this problem. Essentially, you would keep track of the number of good paths that end at each node in the grid. Then, for each node, you would add the number of good paths that end at its adjacent nodes. This would continue until all nodes in the grid have been visited. The final count would be the sum of all the good paths that end at the nodes in the grid.```
```def numOfPaths(n):

# Create a 2D array
# to store results
# of subproblems
dp = [[0 for i in range(n)]
for j in range(n)]

# Base cases
for i in range(n):
dp[0][i] = 1
dp[i][0] = 1

# Fill dp[][] in
# bottom up manner
for i in range(1, n):
for j in range(1, n):

# No. of ways to reach
# cell (i, j) from cell
# (i - 1, j) + (i, j - 1)
# as we can only move
# down and right
dp[i][j] = dp[i - 1][j] + dp[i][j - 1]

# Return result
return dp[n - 1][n - 1]```
```There are several ways to solve this problem. One approach would be to use a breadth-first search algorithm. Another approach would be to use a depth-first search algorithm.

Here is a depth-first search solution:

function numOfGoodPaths(n, m) {
let numPaths = 0;

function dfs(i, j) {
// If we've reached the end of the grid, then we have a good path.
if (i === n - 1 && j === m - 1) {
numPaths++;
return;
}

// Mark the current cell as visited.
const key = `\${i}_\${j}`;
const visited = {};
visited[key] = true;

// Explore the four possible directions from the current cell.
const dirs = [[1, 0], [0, 1], [-1, 0], [0, -1]];
for (let dir of dirs) {
const newI = i + dir[0];
const newJ = j + dir[1];

// Skip over any cells that are out-of-bounds or have already been visited.
if (newI < 0 || newI >= n || newJ < 0 || newJ >= m || visited[`\${newI}_\${newJ}`]) {
continue;
}

dfs(newI, newJ);
}
}

dfs(0, 0);
return numPaths;
}```
```There are several ways to solve this problem. One approach is to use a graph algorithm such as Dijkstra's algorithm or A* search. Another approach is to use a dynamic programming algorithm.

We will use the dynamic programming approach in this solution.

Let dp[i][j] be the number of good paths from node i to node j. Then, we have the following recursive relation:

dp[i][j] = dp[i-1][j] + dp[i][j-1]

where dp[i-1][j] is the number of good paths from node i-1 to node j and dp[i][j-1] is the number of good paths from node i to node j-1.

We can compute dp[i][j] in a bottom-up manner. For example, consider the following grid:

0 1 2 3 4

0 1 2

0 1

0

The base cases are dp[i][0] = dp[0][j] = 1 for all i, j. We can then compute dp[i][j] for i > 0, j > 0 as follows:

dp[1][1] = dp[0][1] + dp[1][0] = 1 + 1 = 2
dp[1][2] = dp[0][2] + dp[1][1] = 1 + 2 = 3
dp[2][1] = dp[1][0] + dp[2][1] = 1 + 1 = 2
dp[2][2] = dp[1][1] + dp[2][1] = 2 + 2 = 4

Thus, the number of good paths from node (2,2) to node (4,4) is 4.```
```There are several ways to approach this problem. One way is to use a breadth-first search algorithm to find all of the possible paths through the grid. Then, you can count the number of paths that pass through each cell and return the total number of paths.

Another way to approach this problem is to use a dynamic programming algorithm. You can create a 2D array to keep track of the number of paths through each cell. Then, you can fill in the array by starting at the top left cell and working your way down and to the right. Each cell will contain the sum of the paths through the cell to the left and the paths through the cell to the top. The total number of paths will be the value in the bottom right cell of the array.```

Scroll to Top
[gravityforms id="5" description="false" titla="false" ajax="true"]