Unique Paths in Grid DP in DSA Typescript

Mental Model

We count how many ways to reach each cell by adding ways from the top and left cells. This builds up the total paths step by step.

Analogy: Imagine walking on a city grid where you can only move right or down. To reach a building, you add the number of ways you could have come from the building above and the building to the left.

Start grid (3x3):
[1] -> [ ] -> [ ]
 ↓      ↓      ↓
[ ] -> [ ] -> [ ]
 ↓      ↓      ↓
[ ] -> [ ] -> [ ]

[1] means starting point with 1 way, empty means unknown ways.

Dry Run Walkthrough

Input: Grid size: 3 rows x 3 columns

Goal: Find total unique paths from top-left corner to bottom-right corner moving only right or down

Step 1: Initialize first row with 1s because only one way to move right along top row

[1] -> [1] -> [1]
 ↓      ↓      ↓
[ ] -> [ ] -> [ ]
 ↓      ↓      ↓
[ ] -> [ ] -> [ ]

Why: Only one way to reach cells in first row: keep moving right

Step 2: Initialize first column with 1s because only one way to move down along left column

[1] -> [1] -> [1]
 ↓      ↓      ↓
[1] -> [ ] -> [ ]
 ↓      ↓      ↓
[1] -> [ ] -> [ ]

Why: Only one way to reach cells in first column: keep moving down

Step 3: Calculate ways for cell (1,1) by adding ways from top (0,1) and left (1,0)

[1] -> [1] -> [1]
 ↓      ↓      ↓
[1] -> [2] -> [ ]
 ↓      ↓      ↓
[1] -> [ ] -> [ ]

Why: Paths to (1,1) = paths from above + paths from left = 1 + 1 = 2

Step 4: Calculate ways for cell (1,2) by adding ways from top (0,2) and left (1,1)

[1] -> [1] -> [1]
 ↓      ↓      ↓
[1] -> [2] -> [3]
 ↓      ↓      ↓
[1] -> [ ] -> [ ]

Why: Paths to (1,2) = 1 (top) + 2 (left) = 3

Step 5: Calculate ways for cell (2,1) by adding ways from top (1,1) and left (2,0)

[1] -> [1] -> [1]
 ↓      ↓      ↓
[1] -> [2] -> [3]
 ↓      ↓      ↓
[1] -> [3] -> [ ]

Why: Paths to (2,1) = 2 (top) + 1 (left) = 3

Step 6: Calculate ways for cell (2,2) by adding ways from top (1,2) and left (2,1)

[1] -> [1] -> [1]
 ↓      ↓      ↓
[1] -> [2] -> [3]
 ↓      ↓      ↓
[1] -> [3] -> [6]

Why: Paths to (2,2) = 3 (top) + 3 (left) = 6

Result:

Final grid with path counts:
[1] -> [1] -> [1]
 ↓      ↓      ↓
[1] -> [2] -> [3]
 ↓      ↓      ↓
[1] -> [3] -> [6]

Total unique paths = 6

Annotated Code

DSA Typescript

class UniquePaths {
  static uniquePaths(m: number, n: number): number {
    if (m === 0 || n === 0) return 0;
    const dp: number[][] = Array.from({ length: m }, () => Array(n).fill(0));

    // Initialize first row with 1s
    for (let col = 0; col < n; col++) {
      dp[0][col] = 1;
    }

    // Initialize first column with 1s
    for (let row = 0; row < m; row++) {
      dp[row][0] = 1;
    }

    // Fill the dp table
    for (let row = 1; row < m; row++) {
      for (let col = 1; col < n; col++) {
        dp[row][col] = dp[row - 1][col] + dp[row][col - 1];
      }
    }

    return dp[m - 1][n - 1];
  }
}

// Driver code
const m = 3;
const n = 3;
const result = UniquePaths.uniquePaths(m, n);
console.log(`Total unique paths in a ${m}x${n} grid: ${result}`);

for (let col = 0; col < n; col++) { dp[0][col] = 1; }

Initialize first row with 1s because only one way to move right

for (let row = 0; row < m; row++) { dp[row][0] = 1; }

Initialize first column with 1s because only one way to move down

for (let row = 1; row < m; row++) { for (let col = 1; col < n; col++) { dp[row][col] = dp[row - 1][col] + dp[row][col - 1]; } }

Calculate paths for each cell by adding ways from top and left cells

return dp[m - 1][n - 1];

Return total unique paths to bottom-right corner

Output

Total unique paths in a 3x3 grid: 6

Complexity Analysis

Time: O(m*n) because we fill each cell in the grid once

Space: O(m*n) because we use a 2D array to store path counts for each cell

vs Alternative: Compared to recursive approach without memoization which is exponential, this DP approach is efficient and avoids repeated calculations

Edge Cases

Grid with 1 row or 1 column

Returns 1 because only one path exists (all right or all down moves)

DSA Typescript

for (let col = 0; col < n; col++) {
  dp[0][col] = 1;
}

Grid size 0 (m=0 or n=0)

Function returns 0 or no paths since grid is invalid or empty

DSA Typescript

Added explicit guard: if (m === 0 || n === 0) return 0;

When to Use This Pattern

When you see a grid and need to count paths moving only right or down, use DP to build solutions from smaller subproblems because each cell depends on top and left neighbors.

Common Mistakes

Mistake: Not initializing first row and first column with 1s
Fix: Explicitly set dp[0][col] and dp[row][0] to 1 before filling the rest of the grid

Mistake: Adding dp[row][col - 1] and dp[row - 1][col] in wrong order or indices
Fix: Ensure dp[row][col] = dp[row - 1][col] + dp[row][col - 1] to correctly sum top and left cells

Summary

Calculates the number of unique paths from top-left to bottom-right in a grid moving only right or down.

Use when counting ways to traverse grids with movement restrictions.

Each cell's paths count is the sum of paths from the cell above and the cell to the left.

Practice

1. What does the problem "Unique Paths in Grid DP" typically ask you to find?

easy

A. The number of ways to move from the top-left to the bottom-right corner of a grid using only right and down moves.

B. The shortest path distance between two points in a grid.

C. The maximum sum path in a grid.

D. The number of obstacles in a grid.

Solution

1. Step 1: Understand the problem statement

   The problem asks for counting all possible paths from the top-left corner to the bottom-right corner of a grid, moving only right or down.

2. Step 2: Identify the key movement constraints

   Only moves to the right or down are allowed, so the problem counts unique sequences of these moves.

3. Final Answer:

   The number of ways to move from the top-left to the bottom-right corner of a grid using only right and down moves. -> Option A

4. Quick Check:

   Unique paths = count of right/down moves sequences [OK]

Hint: Count paths by moving only right or down [OK]

Common Mistakes:

• Confusing unique paths with shortest path length
• Including diagonal moves
• Counting obstacles instead of paths

2. Which of the following TypeScript code snippets correctly initializes a 2D DP array for a grid of size m x n filled with zeros?

easy

A. const dp = Array(m).fill(Array(n).fill(0));

B. const dp = Array(m).map(Array(n).fill(0));

C. const dp = new Array(m).map(new Array(n).fill(0));

D. const dp = Array.from({ length: m }, () => Array(n).fill(0));

Solution

1. Step 1: Understand how to create independent rows

   Using Array.from with a mapping function creates new arrays for each row, avoiding shared references.

2. Step 2: Analyze each option

   const dp = Array(m).fill(Array(n).fill(0)); shares the same inner array reference for all rows, causing bugs. Options C and D pass an array (not a function) to .map(), causing TypeError. const dp = Array.from({ length: m }, () => Array(n).fill(0)); correctly creates independent rows filled with zeros.

3. Final Answer:

   const dp = Array.from({ length: m }, () => Array(n).fill(0)); -> Option D

4. Quick Check:

   Use Array.from with a function for 2D arrays [OK]

Hint: Use Array.from with a function to avoid shared references [OK]

Common Mistakes:

• Using fill with a single array causes shared references
• Passing non-function to .map() causes TypeError
• Not initializing inner arrays separately

3. What is the output of the following TypeScript code for a 3x3 grid?

function uniquePaths(m: number, n: number): number {
  const dp = Array.from({ length: m }, () => Array(n).fill(1));
  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
    }
  }
  return dp[m - 1][n - 1];
}
console.log(uniquePaths(3, 3));

medium

A. 1

B. 3

C. 6

D. 9

Solution

1. Step 1: Initialize dp array with 1s

   Each cell in the first row and first column is 1, representing only one path to those cells.

2. Step 2: Fill dp using sum of top and left cells

   For each cell dp[i][j], sum dp[i-1][j] (top) and dp[i][j-1] (left). For 3x3, dp[2][2] = 6.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Unique paths in 3x3 grid = 6 [OK]

Hint: Sum paths from top and left cells to fill dp [OK]

Common Mistakes:

• Starting loops from 0 instead of 1
• Not initializing first row and column to 1
• Returning wrong dp cell

4. Identify the bug in this TypeScript code for unique paths and select the fix:

function uniquePaths(m: number, n: number): number {
  const dp = Array.from({ length: m }, () => Array(n).fill(0));
  for (let i = 0; i < m; i++) dp[i][0] = 1;
  for (let j = 0; j < n; j++) dp[0][j] = 1;
  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      dp[i][j] = dp[i - 1][j] * dp[i][j - 1];
    }
  }
  return dp[m - 1][n - 1];
}

medium

A. Change multiplication (*) to addition (+) in dp update.

B. Initialize dp with 1s instead of 0s.

C. Start loops from 0 instead of 1.

D. Return dp[0][0] instead of dp[m-1][n-1].

Solution

1. Step 1: Identify the dp update formula

   The code uses multiplication (*) to combine paths, which is incorrect because paths add up, not multiply.

2. Step 2: Correct the dp update to addition

   Replace dp[i - 1][j] * dp[i][j - 1] with dp[i - 1][j] + dp[i][j - 1] to sum paths from top and left.

3. Final Answer:

   Change multiplication (*) to addition (+) in dp update. -> Option A

4. Quick Check:

   Paths combine by addition, not multiplication [OK]

Hint: Sum top and left dp cells, don't multiply [OK]

Common Mistakes:

• Using multiplication instead of addition
• Incorrect dp initialization
• Wrong loop start indices

5. Given a grid with obstacles represented by 1s and empty spaces by 0s, how would you modify the unique paths DP solution to count paths avoiding obstacles?

const grid = [
  [0,0,0],
  [0,1,0],
  [0,0,0]
];

hard

A. Ignore obstacles and calculate paths normally.

B. Set dp[i][j] = 0 if grid[i][j] is 1; else dp[i][j] = dp[i-1][j] + dp[i][j-1].

C. Set dp[i][j] = 1 if grid[i][j] is 1; else 0.

D. Multiply dp[i-1][j] and dp[i][j-1] regardless of obstacles.

Solution

1. Step 1: Mark obstacle cells as unreachable

   If grid[i][j] is 1 (obstacle), set dp[i][j] = 0 to block paths through it.

2. Step 2: Calculate dp for free cells

   For cells without obstacles, dp[i][j] = dp[i-1][j] + dp[i][j-1], summing paths from top and left.

3. Final Answer:

   Set dp[i][j] = 0 if grid[i][j] is 1; else dp[i][j] = dp[i-1][j] + dp[i][j-1]. -> Option B

4. Quick Check:

   Obstacles block paths by setting dp to zero [OK]

Hint: Set dp to zero on obstacles, sum paths otherwise [OK]

Common Mistakes:

• Ignoring obstacles in dp calculation
• Setting dp to 1 on obstacles
• Multiplying paths instead of adding