Unique Paths in Grid DP in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Unique Paths Calculation

What is the output of the following code that calculates unique paths in a 3x3 grid?

DSA Javascript

function uniquePaths(m, n) {
  const dp = Array(m).fill().map(() => 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));

A 9

B 6

C 12

D 3

💡 Hint

Think about how many ways you can move right or down to reach the bottom-right corner.

The number of unique paths in a 3x3 grid is 6 because you can move right or down in different sequences to reach the end.

🧠 Conceptual

intermediate

Understanding DP Table Initialization

Why do we initialize the first row and first column of the DP table with 1 in the Unique Paths problem?

A Because the first row and column represent the maximum number of paths possible.

B Because the first row and column are obstacles and cannot be crossed.

C Because there is only one way to reach any cell in the first row or first column, by moving only right or only down.

D Because the first row and column are not used in the calculation.

💡 Hint

Think about how you can move from the start to cells in the first row or column.

The first row and column can only be reached by moving straight right or straight down respectively, so there is exactly one path to each of those cells.

❓ Predict Output

advanced

Output with Obstacles in Grid

What is the output of the code below that calculates unique paths in a 3x3 grid with an obstacle at position (1,1)?

DSA Javascript

function uniquePathsWithObstacles(obstacleGrid) {
  const m = obstacleGrid.length;
  const n = obstacleGrid[0].length;
  const dp = Array(m).fill().map(() => Array(n).fill(0));
  dp[0][0] = obstacleGrid[0][0] === 0 ? 1 : 0;
  for (let i = 1; i < m; i++) {
    dp[i][0] = (obstacleGrid[i][0] === 0 && dp[i-1][0] === 1) ? 1 : 0;
  }
  for (let j = 1; j < n; j++) {
    dp[0][j] = (obstacleGrid[0][j] === 0 && dp[0][j-1] === 1) ? 1 : 0;
  }
  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      if (obstacleGrid[i][j] === 0) {
        dp[i][j] = dp[i-1][j] + dp[i][j-1];
      } else {
        dp[i][j] = 0;
      }
    }
  }
  return dp[m-1][n-1];
}

const grid = [
  [0,0,0],
  [0,1,0],
  [0,0,0]
];
console.log(uniquePathsWithObstacles(grid));

A 2

B 4

C 3

D 0

💡 Hint

The obstacle blocks some paths, reducing the total count.

With an obstacle in the middle, only two unique paths remain to reach the bottom-right corner.

🔧 Debug

advanced

Identify the Error in DP Implementation

What error will the following code produce when calculating unique paths in a 3x3 grid?

DSA Javascript

function uniquePaths(m, n) {
  const dp = Array(m).fill(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));

A All rows in dp refer to the same array, causing incorrect results.

B Returns 0 because dp is initialized with zeros.

C TypeError because dp is not an array.

D SyntaxError due to missing semicolons.

💡 Hint

Check how the dp array is initialized and whether rows are independent.

Using Array(m).fill(Array(n).fill(1)) makes all rows point to the same array, so updating one row updates all rows, causing wrong results.

🚀 Application

expert

Number of Unique Paths in a 5x7 Grid

How many unique paths are there from the top-left to the bottom-right corner in a 5x7 grid if you can only move right or down?

A 462

B 330

C 792

D 210

💡 Hint

Use the formula for combinations: (m+n-2) choose (m-1).

The number of unique paths is the binomial coefficient C(5+7-2, 5-1) = C(10,4) = 210, but options show 210 as C, so check carefully. Actually, C(10,4) = 210, but the grid is 5x7, so m=5, n=7, total moves = 5-1 + 7-1 = 10 moves, choose 4 downs or 6 rights. C(10,4) = 210. So correct answer is 210 (option D).

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