Why Unique Paths in Grid DP in DSA Javascript?

The Big Idea

What if you could instantly know all possible routes in a city grid without walking them?

The Scenario

Imagine you are in a big city with a grid of streets. You want to walk from your home at the top-left corner to a friend's house at the bottom-right corner. You can only walk right or down. Counting all possible ways by hand is like listing every single route on a huge map.

The Problem

Trying to count all paths manually is slow and confusing. As the grid grows, the number of paths grows very fast, making it impossible to keep track without mistakes. You might forget some routes or count some twice.

The Solution

Dynamic Programming helps by breaking the problem into smaller parts. It remembers how many ways to reach each cell, so you don't count paths again and again. This way, you quickly find the total paths without listing them all.

Before vs After

✗ Before

function countPaths(row, col) {
  if (row === 0 && col === 0) return 1;
  if (row < 0 || col < 0) return 0;
  return countPaths(row - 1, col) + countPaths(row, col - 1);
}

✓ After

function uniquePaths(rows, cols) {
  const dp = Array(rows).fill().map(() => Array(cols).fill(0));
  for (let r = 0; r < rows; r++) dp[r][0] = 1;
  for (let c = 0; c < cols; c++) dp[0][c] = 1;
  for (let r = 1; r < rows; r++) {
    for (let c = 1; c < cols; c++) {
      dp[r][c] = dp[r - 1][c] + dp[r][c - 1];
    }
  }
  return dp[rows - 1][cols - 1];
}

What It Enables

This method lets you quickly find how many ways to reach any point in a grid, even if the grid is very large.

Real Life Example

Delivery drivers use similar calculations to find how many routes they can take through city blocks to reach a destination efficiently.

Key Takeaways

Manual counting of paths is slow and error-prone.

Dynamic Programming stores results to avoid repeated work.

It efficiently calculates total unique paths in a 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