Unique Paths in Grid DP in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the problem statement of Unique Paths in Grid DP?

Find the number of ways to move from the top-left corner to the bottom-right corner of a grid, moving only down or right.

beginner

What does DP stand for and why is it used in Unique Paths?

DP stands for Dynamic Programming. It is used to store results of subproblems (number of ways to reach each cell) to avoid repeated calculations.

beginner

How do you initialize the first row and first column in the DP table for Unique Paths?

Set all cells in the first row and first column to 1 because there is only one way to reach cells in the first row (all right moves) or first column (all down moves).

intermediate

What is the recurrence relation used in Unique Paths DP?

dp[i][j] = dp[i-1][j] + dp[i][j-1], meaning ways to reach cell (i,j) is sum of ways from the cell above and the cell to the left.

beginner

Why can't we move diagonally in the Unique Paths problem?

The problem constraints allow only moves to the right or down, so diagonal moves are not allowed and not counted.

In Unique Paths, what does dp[i][j] represent?

A Number of ways to reach cell (i,j)

B Minimum steps to reach cell (i,j)

C Maximum value in row i

D Number of obstacles in cell (i,j)

dp[i][j] stores the number of unique ways to reach the cell at row i and column j.

How many ways are there to reach any cell in the first row?

A 1

B 0

C Depends on the column

D Equal to row number

Only one way: moving right from the start.

Which moves are allowed in the Unique Paths problem?

A Up and left

B All directions

C Right and down

D Diagonal only

Only moves to the right or down are allowed.

What is the base case for dp[0][0] in Unique Paths?

A 0

B 1

C Depends on grid size

D Undefined

Starting cell has 1 way to be at itself.

If dp[i][j] = dp[i-1][j] + dp[i][j-1], what does this mean?

A Sum of all cells in the grid

B Ways to reach (i,j) come from below and right cells

C Number of obstacles in row i and column j

D Ways to reach (i,j) come from above and left cells

You can only come from the cell above or the cell to the left.

Explain how to use dynamic programming to find the number of unique paths in a grid.

Think about how you can build the solution step by step from the start.

Describe why the first row and first column in the DP table are filled with 1s in the Unique Paths problem.

Imagine walking only along the edges of the grid.

Practice

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

easy

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

B. The shortest path length from top-left to bottom-right in a grid.

C. The maximum sum path from top-left to bottom-right in a grid.

D. The number of obstacles in the grid.

Solution

1. Step 1: Understand the problem goal

   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: Differentiate from other path problems

   It is not about shortest path length, maximum sum, or obstacles count, but about counting unique paths.

3. Final Answer:

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

4. Quick Check:

   Unique paths count = number of ways to move right/down [OK]

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

Common Mistakes:

• Confusing unique paths with shortest path length
• Thinking about obstacles when none are given
• Counting diagonal moves

2. Which of the following JavaScript code snippets correctly initializes a 2D array dp of size m x n filled with zeros for the Unique Paths problem?

easy

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

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

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

D. const dp = Array(m).map(() => Array(n));

Solution

1. Step 1: Understand array initialization in JavaScript

   Using Array.from with a mapping function creates independent inner arrays, 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 create inner arrays with undefined elements instead of zeros (missing .fill(0)). const dp = Array.from({ length: m }, () => Array(n).fill(0)); correctly creates independent arrays filled with zeros.

3. Final Answer:

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

4. Quick Check:

   Use Array.from with arrow function for 2D array [OK]

Hint: Use Array.from with function to avoid shared inner arrays [OK]

Common Mistakes:

• Using fill with array directly causing shared references
• Not filling inner arrays with zeros (elements undefined)
• Not initializing inner arrays separately

3. What is the output of the following JavaScript code for uniquePaths(3, 3)?

function uniquePaths(m, n) {
  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];
}
console.log(uniquePaths(3, 3));

medium

A. 3

B. 6

C. 9

D. 12

Solution

1. Step 1: Initialize dp array and base cases

   Set first row and first column to 1 because only one way to reach cells in first row or column.

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

   Calculate dp[1][1] = 1+1=2, dp[1][2] = 1+2=3, dp[2][1] = 2+1=3, dp[2][2] = 3+3=6.

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Unique paths in 3x3 grid = 6 [OK]

Hint: Sum paths from top and left neighbors [OK]

Common Mistakes:

• Not initializing first row and column to 1
• Adding instead of summing dp[i-1][j] and dp[i][j-1]
• Off-by-one errors in loops

4. The following code attempts to solve the Unique Paths problem but overwrites values across multiple rows unexpectedly for uniquePaths(3, 3). What is the bug?

function uniquePaths(m, n) {
  const dp = Array(m).fill(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. The dp array is not initialized with zeros.

B. The loops start from 1 instead of 0.

C. Using fill with array causes all rows to reference the same array.

D. The return statement is outside the function.

Solution

1. Step 1: Analyze dp initialization

   Using Array(m).fill(Array(n).fill(0)) makes all rows point to the same inner array, so updating one row updates all rows.

2. Step 2: Understand effect on dp updates

   Setting dp[i][0] = 1 changes the same array for all rows. During table filling, updates overwrite values from earlier rows (e.g., dp[1][1] becomes 3 instead of 2), causing incorrect intermediate values.

3. Final Answer:

   Using fill with array causes all rows to reference the same array. -> Option C

4. Quick Check:

   Shared references cause wrong dp updates [OK]

Hint: Avoid fill with array; use Array.from for independent rows [OK]

Common Mistakes:

• Using fill with array causing shared references
• Incorrect loop ranges
• Misplaced return statement

5. Given a grid with obstacles represented by a 2D array obstacleGrid where 1 means obstacle and 0 means free cell, how would you modify the Unique Paths DP solution to count unique paths avoiding obstacles?

hard

A. Set dp[i][j] = 1 for all cells without checking obstacles.

B. Ignore obstacles and calculate paths as usual.

C. Use recursion without memoization to count paths.

D. Initialize dp with zeros, set dp[0][0] = 1 if no obstacle, then for each cell dp[i][j] = 0 if obstacle else dp[i-1][j] + dp[i][j-1].

Solution

1. Step 1: Initialize dp considering obstacles

   Set dp[0][0] to 1 only if the start cell has no obstacle; otherwise 0.

2. Step 2: Update dp cells with obstacle check

   For each cell, if obstacleGrid[i][j] is 1, dp[i][j] = 0 (no path). Else dp[i][j] = dp[i-1][j] + dp[i][j-1].

3. Step 3: Return dp[m-1][n-1]

   This gives the count of unique paths avoiding obstacles.

4. Final Answer:

   Initialize dp with zeros, set dp[0][0] = 1 if no obstacle, then for each cell dp[i][j] = 0 if obstacle else dp[i-1][j] + dp[i][j-1]. -> Option D

5. Quick Check:

   Obstacle cells block paths by setting dp to 0 [OK]

Hint: Set dp=0 for obstacles, sum paths only for free cells [OK]

Common Mistakes:

• Ignoring obstacles in dp calculation
• Not initializing start cell properly
• Using recursion without memoization causing inefficiency