Unique Paths in Grid DP in DSA Typescript - Execution Trace

Concept Flow - Unique Paths in Grid DP

Start at top-left cell

↓

Initialize DP table with 1s in first row and column

↓

For each cell (i,j) from (1,1) to (m-1,n-1)

↓

Calculate dp[i

↓

Repeat until bottom-right cell

↓

Return dp[m-1

We fill a grid table where each cell counts paths from start by adding paths from top and left cells, starting with 1s on the first row and column.

Execution Sample

DSA Typescript

function uniquePaths(m: number, n: number): number {
  const dp = Array.from({length: m}, (_, i) => Array.from({length: n}, (_, j) => (i === 0 || j === 0) ? 1 : 0));
  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];
}

This code calculates the number of unique paths in an m x n grid moving only right or down.

Execution Table

Step	Operation	Cell Updated	Value Computed	DP Table State	Explanation
0	Initialize DP table	All cells	1 in first row and column, 0 elsewhere	[[1,1,1],[1,0,0],[1,0,0]]	First row and column set to 1 because only one way to reach those cells
1	Calculate dp[1][1]	(1,1)	dp[0][1] + dp[1][0] = 1 + 1 = 2	[[1,1,1],[1,2,0],[1,0,0]]	Paths to (1,1) come from top and left cells
2	Calculate dp[1][2]	(1,2)	dp[0][2] + dp[1][1] = 1 + 2 = 3	[[1,1,1],[1,2,3],[1,0,0]]	Paths to (1,2) sum top and left paths
3	Calculate dp[2][1]	(2,1)	dp[1][1] + dp[2][0] = 2 + 1 = 3	[[1,1,1],[1,2,3],[1,3,0]]	Paths to (2,1) sum top and left paths
4	Calculate dp[2][2]	(2,2)	dp[1][2] + dp[2][1] = 3 + 3 = 6	[[1,1,1],[1,2,3],[1,3,6]]	Paths to bottom-right cell is total unique paths
5	Return result	dp[2][2]	6	[[1,1,1],[1,2,3],[1,3,6]]	Total unique paths in 3x3 grid is 6

💡 All cells computed, bottom-right cell dp[m-1][n-1] returned as result

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	Final
dp	[[1,1,1],[1,0,0],[1,0,0]]	[[1,1,1],[1,2,0],[1,0,0]]	[[1,1,1],[1,2,3],[1,0,0]]	[[1,1,1],[1,2,3],[1,3,0]]	[[1,1,1],[1,2,3],[1,3,6]]	[[1,1,1],[1,2,3],[1,3,6]]

Key Moments - 3 Insights

Why do we set the first row and first column cells to 1 initially?

Because from the start cell, you can only move right along the first row or down along the first column, so there is exactly one way to reach each of those cells. This is shown in execution_table row 0.

Why do we add dp[i-1][j] and dp[i][j-1] to get dp[i][j]?

Because to reach cell (i,j), you can only come from the cell above (i-1,j) or the cell to the left (i,j-1). So total paths to (i,j) is sum of paths to those two cells, as shown in steps 1 to 4 in execution_table.

What does dp[m-1][n-1] represent at the end?

It represents the total number of unique paths from the top-left corner to the bottom-right corner of the grid, as shown in step 5 of execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at Step 3. What is the value computed for cell (2,1)?

A 4

B 2

C 3

D 6

💡 Hint

Check the 'Value Computed' column in execution_table row 3.

At which step does the dp table first show a value greater than 3?

A Step 4

B Step 2

C Step 3

D Step 1

💡 Hint

Look at the 'Value Computed' column and the DP Table State in execution_table rows.

If the grid size was 2x2 instead of 3x3, what would dp[1][1] be after computation?

A 1

B 2

C 3

D 4

💡 Hint

Recall dp[i][j] = dp[i-1][j] + dp[i][j-1], and first row and column are 1.

Concept Snapshot

Unique Paths in Grid DP:
- Use a 2D dp array of size m x n
- Initialize first row and column with 1
- For each cell dp[i][j], dp[i][j] = dp[i-1][j] + dp[i][j-1]
- Result is dp[m-1][n-1]
- Counts unique ways moving only right or down

Full Transcript

This concept shows how to find the number of unique paths in a grid from the top-left to bottom-right corner. We create a 2D table where each cell stores the count of ways to reach it. The first row and column are set to 1 because you can only move right or down along edges. For each other cell, the number of ways is the sum of ways from the cell above and the cell to the left. We fill the table row by row until we reach the bottom-right cell, which holds the total unique paths. The example uses a 3x3 grid and shows step-by-step how the dp table fills up, ending with 6 unique paths.

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