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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Unique Paths Calculation

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

DSA Go

package main
import "fmt"
func uniquePaths(m int, 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]
}
func main() {
    fmt.Println(uniquePaths(3, 3))
}

A 5

B 3

C 6

D 9

💡 Hint

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

The code uses dynamic programming to count paths. For a 3x3 grid, the number of unique paths is 6.

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

B Because the first row and column represent obstacles that block paths.

C Because the first row and column are always zero in the DP table.

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

💡 Hint

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

You can only move right to reach cells in the first row and only down to reach cells in the first column, so each has exactly one path.

🔧 Debug

advanced

Identify the Bug in DP Update Loop

What error will occur when running this code snippet for unique paths in a grid?

DSA Go

func uniquePaths(m int, 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]
}

A no error, returns correct result

B compilation error: missing return statement

C returns 0 instead of correct paths

D runtime error: index out of range

💡 Hint

Check the loop boundaries carefully.

The loops use <= which causes index to go out of bounds since valid indices are 0 to m-1 and 0 to n-1.

🚀 Application

advanced

Number of Unique Paths in a 4x5 Grid

Using the unique paths DP approach, how many unique paths are there in a 4 rows by 5 columns grid?

A 56

B 35

C 70

D 20

💡 Hint

Use the formula or DP logic: paths = combination(m+n-2, m-1).

Number of unique paths = C(4+5-2, 4-1) = C(7,3) = 35. But careful: 4 rows and 5 columns means m=4, n=5, so paths = C(7,3) = 35. The correct answer is 35.

❓ Predict Output

expert

Output of Unique Paths with Obstacles

What is the output of this Go code that calculates unique paths in a grid with obstacles? 0 means free cell, 1 means obstacle.

DSA Go

package main
import "fmt"
func uniquePathsWithObstacles(obstacleGrid [][]int) int {
    m := len(obstacleGrid)
    n := len(obstacleGrid[0])
    dp := make([][]int, m)
    for i := range dp {
        dp[i] = make([]int, n)
    }
    if obstacleGrid[0][0] == 1 {
        return 0
    }
    dp[0][0] = 1
    for i := 1; i < m; i++ {
        if obstacleGrid[i][0] == 0 && dp[i-1][0] == 1 {
            dp[i][0] = 1
        }
    }
    for j := 1; j < n; j++ {
        if obstacleGrid[0][j] == 0 && dp[0][j-1] == 1 {
            dp[0][j] = 1
        }
    }
    for i := 1; i < m; i++ {
        for j := 1; j < n; j++ {
            if obstacleGrid[i][j] == 0 {
                dp[i][j] = dp[i-1][j] + dp[i][j-1]
            }
        }
    }
    return dp[m-1][n-1]
}
func main() {
    grid := [][]int{
        {0,0,0},
        {0,1,0},
        {0,0,0},
    }
    fmt.Println(uniquePathsWithObstacles(grid))
}

A 2

B 4

C 3

D 1

💡 Hint

Obstacles block paths. Count only valid routes avoiding obstacles.

The obstacle at position (1,1) blocks some paths. The total unique paths from top-left to bottom-right is 2.

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