Unique Paths in Grid DP in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the first row of the DP table with 1s.

DSA C

for (int j = 0; j < n; j++) {
    dp[0][j] = [1];
}

Drag options to blanks, or click blank then click option'

Aj

B0

C1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[0][j] to 0 instead of 1.

Using the loop variable j as the value.

The first row has only one way to reach each cell by moving right, so dp[0][j] should be set to 1.

2fill in blank

medium

Complete the code to initialize the first column of the DP table with 1s.

DSA C

for (int i = 0; i < m; i++) {
    dp[i][0] = [1];
}

Drag options to blanks, or click blank then click option'

Ai

Bm

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[i][0] to 0 instead of 1.

Using the loop variable i as the value.

The first column has only one way to reach each cell by moving down, so dp[i][0] should be set to 1.

3fill in blank

hard

Fix the error in the code to correctly compute the number of unique paths for each cell.

DSA C

for (int i = 1; i < m; i++) {
    for (int j = 1; j < n; j++) {
        dp[i][j] = dp[i - 1][j] [1] dp[i][j - 1];
    }
}

Drag options to blanks, or click blank then click option'

A+

B*

C/

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

Using division which is not meaningful here.

The number of unique paths to a cell is the sum of paths from the cell above and the cell to the left, so use '+'.

4fill in blank

hard

Fill both blanks to complete the nested loops iterating over the grid cells starting from (1,1).

DSA C

for (int [1] = 1; [2] < m; [1]++) {
    for (int j = 1; j < n; j++) {
        dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
    }
}

Drag options to blanks, or click blank then click option'

Ai

Bj

Ck

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables in declaration and condition.

Using 'j' for the outer loop which is for columns.

The outer loop variable should be 'i' to iterate rows from 1 to m-1.

5fill in blank

hard

Fill all three blanks to return the total unique paths from top-left to bottom-right in the grid.

DSA C

int uniquePaths(int m, int n) {
    int dp[m][n];
    for (int i = 0; i < m; i++) dp[i][0] = 1;
    for (int j = 0; j < n; j++) dp[0][j] = 1;
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
            dp[i][j] = dp[i - 1][j] [1] dp[i][j - 1];
        }
    }
    return dp[[2]][[3]];
}

Drag options to blanks, or click blank then click option'

A+

B-

Cm - 1

Dn - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' instead of '+' for summing paths.

Returning dp[m][n] which is out of bounds.

Use '+' to sum paths from top and left. Return dp[m-1][n-1] for bottom-right cell.

Practice

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

easy

A. The maximum sum path from the top-left to the bottom-right corner of a grid.

B. The shortest path length from the top-left to the bottom-right corner of a grid.

C. The number of obstacles in the grid.

D. The number of different ways to move from the top-left to the bottom-right corner of a grid using only right and down moves.

Solution

1. Step 1: Understand the problem goal

   The problem asks for counting all possible paths from the start (top-left) to the end (bottom-right) moving only right or down.

2. Step 2: Clarify movement constraints

   Only moves allowed are right or down, no diagonal or backward moves.

3. Final Answer:

   The number of different ways to move from the top-left to the bottom-right corner of a grid using only right and down moves. -> Option D

4. Quick Check:

   Unique paths = count of right/down moves combinations [OK]

Hint: Count paths by moving only right or down [OK]

Common Mistakes:

• Thinking diagonal moves are allowed
• Confusing path count with shortest path
• Counting obstacles instead of paths

2. Which of the following is the correct way to initialize the DP array for a grid of size m x n in C?

easy

A. int dp[m][n] = {0};

B. int dp[m][n]; dp = 0;

C. int dp[m][n]; memset(dp, 0, sizeof(dp));

D. int dp[m][n] = NULL;

Solution

1. Step 1: Understand array initialization in C

   Arrays cannot be assigned directly after declaration; memset is used to set all bytes to zero.

2. Step 2: Check each option

   int dp[m][n] = {0}; tries to initialize with {0} which works only for static arrays, not variable length arrays. int dp[m][n]; memset(dp, 0, sizeof(dp)); uses memset correctly. int dp[m][n]; dp = 0; tries invalid assignment. int dp[m][n] = NULL; assigns NULL which is invalid for arrays.

3. Final Answer:

   int dp[m][n]; memset(dp, 0, sizeof(dp)); -> Option C

4. Quick Check:

   Use memset to zero initialize arrays in C [OK]

Hint: Use memset to zero initialize arrays in C [OK]

Common Mistakes:

• Trying to assign 0 directly to array
• Using NULL for arrays
• Assuming {0} works for variable length arrays

3. What is the output of the following code snippet for m=3, n=3?

int uniquePaths(int m, int n) {
    int dp[m][n];
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < m; i++) dp[i][0] = 1;
    for (int j = 0; j < n; j++) dp[0][j] = 1;
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
            dp[i][j] = dp[i-1][j] + dp[i][j-1];
        }
    }
    return dp[m-1][n-1];
}
printf("%d", uniquePaths(3, 3));

medium

A. 3

B. 6

C. 9

D. 1

Solution

1. Step 1: Initialize first row and column

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

2. Step 2: Fill dp table using sum of top and left cells

   For each cell dp[i][j], value = dp[i-1][j] + dp[i][j-1]. For 3x3 grid, dp[2][2] = 6.

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Unique paths in 3x3 grid = 6 [OK]

Hint: Sum paths from top and left cells to fill DP [OK]

Common Mistakes:

• Not initializing first row and column to 1
• Adding instead of summing dp[i-1][j] and dp[i][j-1]
• Returning wrong dp cell

4. Find the error in this code snippet for unique paths calculation:

int uniquePaths(int m, int n) {
    int dp[m][n];
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i <= m; i++) dp[i][0] = 1;
    for (int j = 0; j <= n; j++) dp[0][j] = 1;
    for (int i = 1; i < m; i++) {
        for (int 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 loops initializing dp go out of bounds causing runtime error.

B. The dp array is not zero initialized.

C. The return statement returns wrong cell.

D. The nested loops start from 0 instead of 1.

Solution

1. Step 1: Check loop bounds for initialization

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

2. Step 2: Identify correct loop bounds

   Loops should use < m and < n to avoid accessing invalid indices.

3. Final Answer:

   The loops initializing dp go out of bounds causing runtime error. -> Option A

4. Quick Check:

   Array indices must be within 0 to size-1 [OK]

Hint: Use < not <= for array indices in loops [OK]

Common Mistakes:

• Using <= instead of < in loops
• Ignoring array bounds causing crashes
• Incorrect initialization of dp array

5. Given a 4x4 grid with some blocked cells (obstacles) represented by 1, how would you modify the unique paths DP solution to count paths avoiding 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 cell is blocked; else dp[i][j] = dp[i-1][j] + dp[i][j-1].

B. Ignore obstacles and calculate paths normally.

C. Set dp[i][j] = 1 for blocked cells and 0 otherwise.

D. Only count paths that move diagonally to avoid obstacles.

Solution

1. Step 1: Mark blocked cells in dp as zero

   If a cell is blocked (value 1), no paths can go through it, so dp[i][j] = 0.

2. Step 2: Calculate dp for free cells

   For free cells (value 0), dp[i][j] = dp[i-1][j] + dp[i][j-1], summing paths from top and left.

3. Final Answer:

   Set dp[i][j] = 0 if cell is blocked; else dp[i][j] = dp[i-1][j] + dp[i][j-1]. -> Option A

4. Quick Check:

   Blocked cells have zero paths [OK]

Hint: Zero dp for obstacles, sum paths for free cells [OK]

Common Mistakes:

• Ignoring obstacles in path count
• Assigning 1 to blocked cells
• Using diagonal moves to avoid obstacles