Unique Paths in Grid DP in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the problem statement of Unique Paths in Grid DP?

Find the number of ways to move from the top-left corner to the bottom-right corner of a grid, moving only right or down.

beginner

What does the DP state represent in Unique Paths problem?

DP[i][j] represents the number of unique paths to reach cell (i, j) from the start.

intermediate

What is the recurrence relation used in Unique Paths DP?

DP[i][j] = DP[i-1][j] + DP[i][j-1], summing paths from the top and left cells.

beginner

Why do we initialize the first row and first column with 1 in Unique Paths DP?

Because there is only one way to reach any cell in the first row (all moves right) or first column (all moves down).

intermediate

How does Unique Paths DP handle obstacles in the grid?

Cells with obstacles have 0 paths; DP skips adding paths through those cells.

In Unique Paths DP, what does DP[i][j] represent?

A Maximum value in row i

B Minimum steps to cell (i, j)

C Number of unique paths to cell (i, j)

D Sum of all cells in column j

DP[i][j] stores the count of unique ways to reach cell (i, j).

What is the base case for the first row in Unique Paths DP?

A All zeros

B Decreasing sequence

C Increasing sequence

D All ones

First row cells have 1 path each, moving only right.

Which moves are allowed in Unique Paths problem?

A Down and right

B Up and right

C Up and left

D Down and left

Only moves down or right are allowed.

How do obstacles affect the DP table in Unique Paths with obstacles?

A They double the paths

B They set paths to zero for that cell

C They have no effect

D They add one extra path

Obstacles block paths, so DP value is zero for those cells.

What is the time complexity of Unique Paths DP solution?

A O(m*n)

B O(m+n)

C O(m^2*n^2)

D O(1)

We fill an m by n DP table once, so O(m*n).

Explain how to build the DP table for Unique Paths problem step-by-step.

Think about how you move only right or down and how paths add up.

Describe how obstacles change the Unique Paths DP approach.

Obstacles block movement, so no paths go through them.

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