Why Unique Paths in Grid DP in DSA Go?

The Big Idea

What if you could instantly know all possible ways to reach your destination without walking them?

The Scenario

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

The Problem

Manually counting each path 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 miss some paths or count some twice.

The Solution

Dynamic Programming (DP) helps by breaking the problem into smaller parts. It remembers how many ways there are to reach each cell, so you don't have to count paths repeatedly. This saves time and avoids errors.

Before vs After

✗ Before

func countPaths(row, col int) int {
    if row == 0 && col == 0 {
        return 1
    }
    ways := 0
    if row > 0 {
        ways += countPaths(row-1, col)
    }
    if col > 0 {
        ways += countPaths(row, col-1)
    }
    return ways
}

✓ After

func uniquePaths(rows, cols int) int {
    dp := make([][]int, rows)
    for i := range dp {
        dp[i] = make([]int, cols)
    }
    for i := 0; i < rows; i++ {
        dp[i][0] = 1
    }
    for j := 0; j < cols; j++ {
        dp[0][j] = 1
    }
    for i := 1; i < rows; i++ {
        for j := 1; j < cols; j++ {
            dp[i][j] = dp[i-1][j] + dp[i][j-1]
        }
    }
    return dp[rows-1][cols-1]
}

What It Enables

It lets you quickly find the number of ways to reach any point in a grid without getting lost in counting.

Real Life Example

Delivery drivers use similar calculations to find how many routes they can take to deliver packages in a city with grid-like streets.

Key Takeaways

Manual counting is slow and error-prone for large grids.

Dynamic Programming stores intermediate results to avoid repeated work.

This method efficiently calculates total unique paths in a grid.

Practice

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

easy

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

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

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 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 meaning of "Unique Paths"

   Unique paths means counting distinct ways, not shortest or maximum sum paths.

3. Final Answer:

   The number of ways to move from top-left to bottom-right in a grid using only right and down moves -> Option B

4. Quick Check:

   Unique Paths = Count of right/down moves paths [OK]

Hint: Unique Paths count all right/down routes from start to end [OK]

Common Mistakes:

• Confusing unique paths with shortest path length
• Thinking about obstacles when none are given
• Mixing path count with path sum

2. Which of the following Go code snippets correctly initializes a 2D slice dp of size m x n with zeros for Unique Paths DP?

easy

A. dp := [][]int{}
dp = append(dp, make([]int, m))
dp = append(dp, make([]int, n))

B. dp := make([]int, m*n)
for i := 0; i < m*n; i++ {
dp[i] = 0
}

C. var dp [m][n]int
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
dp[i][j] = 1
}
}

D. dp := make([][]int, m)
for i := range dp {
dp[i] = make([]int, n)
}

Solution

1. Step 1: Understand Go slice initialization for 2D arrays

   In Go, a 2D slice is created by first making the outer slice, then initializing each inner slice.

2. Step 2: Check each option

   dp := make([][]int, m) for i := range dp { dp[i] = make([]int, n) } correctly creates a slice of length m, then each inner slice of length n, all initialized to zero by default.

3. Final Answer:

   dp := make([][]int, m) for i := range dp { dp[i] = make([]int, n) } -> Option D

4. Quick Check:

   2D slice init with nested make loops = dp := make([][]int, m) for i := range dp { dp[i] = make([]int, n) } [OK]

Hint: Use nested make to create 2D slices in Go [OK]

Common Mistakes:

• Using one-dimensional slice for 2D data
• Initializing inner slices with wrong length
• Appending slices incorrectly

3. What is the output of the following Go code for Unique Paths in a 3x3 grid?

func uniquePaths(m, n int) int {
    dp := make([][]int, m)
    for i := range dp {
        dp[i] = make([]int, n)
    }
    for i := 0; i < m; i++ {
        dp[i][0] = 1
    }
    for j := 0; j < n; j++ {
        dp[0][j] = 1
    }
    for i := 1; i < m; i++ {
        for j := 1; j < n; j++ {
            dp[i][j] = dp[i-1][j] + dp[i][j-1]
        }
    }
    return dp[m-1][n-1]
}

fmt.Println(uniquePaths(3, 3))

medium

A. 3

B. 9

C. 6

D. 1

Solution

1. Step 1: Initialize dp with 1s on first row and column

   First row and first column are set to 1 because only one way to reach those cells (all right or all down).

2. Step 2: Fill dp table using dp[i][j] = dp[i-1][j] + dp[i][j-1]

   Calculate paths for each cell by summing paths from top and left cells. For 3x3 grid, dp[2][2] = 6.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Unique paths in 3x3 grid = 6 [OK]

Hint: Use dp[i][j] = dp[i-1][j] + dp[i][j-1] to fill grid [OK]

Common Mistakes:

• Not initializing first row and column to 1
• Using wrong indices in dp update
• Returning wrong dp cell

4. The following Go code for Unique Paths has a bug. What is the error?

func uniquePaths(m, n int) int {
    dp := make([][]int, m)
    for i := range dp {
        dp[i] = make([]int, n)
    }
    for i := 0; i < m; i++ {
        dp[i][0] = 1
    }
    for j := 0; j < n; j++ {
        dp[0][j] = 1
    }
    for i := 1; i < m; i++ {
        for 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 update uses multiplication instead of addition

B. The dp slice is not initialized properly

C. The loops for initializing first row and column are swapped

D. The function returns dp[0][0] instead of dp[m-1][n-1]

Solution

1. Step 1: Check dp update formula

   The code uses dp[i][j] = dp[i-1][j] * dp[i][j-1], which multiplies paths instead of adding them.

2. Step 2: Correct formula for Unique Paths

   The correct formula is dp[i][j] = dp[i-1][j] + dp[i][j-1] to sum the number of ways from top and left cells.

3. Final Answer:

   The dp update uses multiplication instead of addition -> Option A

4. Quick Check:

   Use addition, not multiplication in dp update [OK]

Hint: Sum paths from top and left, don't multiply [OK]

Common Mistakes:

• Using * instead of + in dp update
• Forgetting to initialize first row/column
• Returning wrong dp cell

5. Given a 4x4 grid with some obstacles represented by 1s, how would you modify the Unique Paths DP approach to count paths avoiding obstacles?
Obstacles grid:

0 0 0 0
0 1 0 0
0 0 1 0
0 0 0 0

hard

A. Set dp[i][j] = 0 if obstacle at (i,j), else dp[i][j] = dp[i-1][j] + dp[i][j-1]

B. Ignore obstacles and use normal dp formula

C. Set dp[i][j] = 1 if obstacle at (i,j), else 0

D. Use multiplication of dp[i-1][j] and dp[i][j-1] for obstacles

Solution

1. Step 1: Identify obstacle handling in DP

   When a cell has an obstacle, no path can go through it, so dp[i][j] must be 0.

2. Step 2: Modify dp update to skip obstacles

   For cells without obstacles, dp[i][j] = dp[i-1][j] + dp[i][j-1]. For obstacles, dp[i][j] = 0.

3. Final Answer:

   Set dp[i][j] = 0 if obstacle at (i,j), else dp[i][j] = dp[i-1][j] + dp[i][j-1] -> Option A

4. Quick Check:

   Obstacles block paths, set dp to zero there [OK]

Hint: Zero dp at obstacles, sum paths otherwise [OK]

Common Mistakes:

• Ignoring obstacles in dp calculation
• Setting dp to 1 at obstacles
• Multiplying dp values instead of adding