Unique Paths in Grid DP in DSA Go - Build from Scratch

Unique Paths in Grid DP

📖 Scenario: Imagine you are helping a robot find how many different ways it can move from the top-left corner to the bottom-right corner of a grid. The robot can only move right or down.

🎯 Goal: Build a program that calculates the number of unique paths the robot can take in a grid of size m x n using dynamic programming.

📋 What You'll Learn

Create two integer variables m and n for grid dimensions

Create a 2D slice dp to store the number of ways to reach each cell

Use nested for loops to fill the dp slice based on previous results

Print the number of unique paths from top-left to bottom-right

💡 Why This Matters

🌍 Real World

Robots navigating factory floors or delivery drones planning routes in a grid-like city.

💼 Career

Understanding dynamic programming helps solve optimization problems in software engineering and algorithm design.

1

Set grid dimensions

Create two integer variables called m and n and set m = 3 and n = 3 to represent a 3 by 3 grid.

DSA Go

# Your code here

Hint

Use m := 3 and n := 3 to create the grid size variables.

2

Create the DP table

Create a 2D slice of integers called dp with m rows and n columns. Initialize all values to 1.

DSA Go

package main

func main() {
    m := 3
    n := 3
    // Create dp slice with m rows and n columns, all set to 1
    // Your code here
}

Hint

Use make([][]int, m) to create rows and a nested loop to fill each row with 1s.

3

Fill the DP table with unique paths counts

Use nested for loops with variables i and j starting from 1 to update dp[i][j] as the sum of dp[i-1][j] and dp[i][j-1].

DSA Go

package main

func main() {
    m := 3
    n := 3
    dp := make([][]int, m)
    for i := 0; i < m; i++ {
        dp[i] = make([]int, n)
        for j := 0; j < n; j++ {
            dp[i][j] = 1
        }
    }
    // Update dp values starting from row 1 and column 1
    // Your code here
}

Hint

Start loops from 1 because the first row and column are already set to 1.

4

Print the number of unique paths

Use fmt.Println to print the value of dp[m-1][n-1], which is the number of unique paths to the bottom-right corner.

DSA Go

package main

import "fmt"

func main() {
    m := 3
    n := 3
    dp := make([][]int, m)
    for i := 0; i < m; i++ {
        dp[i] = make([]int, n)
        for j := 0; j < n; j++ {
            dp[i][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]
        }
    }
    // Print the number of unique paths
    // Your code here
}

Hint

Use fmt.Println(dp[m-1][n-1]) to print the final count.

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