Unique Paths in Grid DP in DSA C - Execution Trace

Concept Flow - Unique Paths in Grid DP

Start at top-left cell (0,0)

↓

Initialize DP table with 1s in first row and first column

↓

For each cell (i,j) from (1,1) to (m-1,n-1)

↓

Calculate dp[i

↓

Repeat until bottom-right cell (m-1,n-1)

↓

Return dp[m-1

We fill a table where each cell counts paths from start by adding paths from top and left cells, building up to the bottom-right.

Execution Sample

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] + dp[i][j-1];
    }
  }
  return dp[m-1][n-1];
}

This code counts unique paths in a grid by dynamic programming, summing paths from above and left cells.

Execution Table

Step	Operation	Cell Updated	Value Computed	DP Table State
1	Initialize first column	(0,0)	1	[[1, _, _], [_, _, _], [_, _, _]]
2	Initialize first column	(1,0)	1	[[1, _, _], [1, _, _], [_, _, _]]
3	Initialize first column	(2,0)	1	[[1, _, _], [1, _, _], [1, _, _]]
4	Initialize first row	(0,0)	1	[[1, _, _], [1, _, _], [1, _, _]]
5	Initialize first row	(0,1)	1	[[1, 1, _], [1, _, _], [1, _, _]]
6	Initialize first row	(0,2)	1	[[1, 1, 1], [1, _, _], [1, _, _]]
7	Compute dp[1][1]	(1,1)	dp[0][1] + dp[1][0] = 1 + 1 = 2	[[1, 1, 1], [1, 2, _], [1, _, _]]
8	Compute dp[1][2]	(1,2)	dp[0][2] + dp[1][1] = 1 + 2 = 3	[[1, 1, 1], [1, 2, 3], [1, _, _]]
9	Compute dp[2][1]	(2,1)	dp[1][1] + dp[2][0] = 2 + 1 = 3	[[1, 1, 1], [1, 2, 3], [1, 3, _]]
10	Compute dp[2][2]	(2,2)	dp[1][2] + dp[2][1] = 3 + 3 = 6	[[1, 1, 1], [1, 2, 3], [1, 3, 6]]
11	Return result	dp[2][2]	6	[[1, 1, 1], [1, 2, 3], [1, 3, 6]]

💡 All cells computed, bottom-right cell dp[2][2] holds total unique paths.

Variable Tracker

Variable	Start	After Step 7	After Step 8	After Step 9	After Step 10	Final
dp[0][0]	undefined	1	1	1	1	1
dp[1][1]	undefined	2	2	2	2	2
dp[1][2]	undefined	undefined	3	3	3	3
dp[2][1]	undefined	undefined	undefined	3	3	3
dp[2][2]	undefined	undefined	undefined	undefined	6	6

Key Moments - 3 Insights

Why do we set the first row and first column cells to 1 before filling the rest?

Because from the start cell, you can only move right along the first row or down along the first column, so there is exactly 1 unique path to each of these cells. This is shown in steps 1-6 in the execution_table.

Why do we add dp[i-1][j] and dp[i][j-1] to get dp[i][j]?

Because any path to cell (i,j) must come from either the cell above (i-1,j) or the cell to the left (i,j-1). So total paths to (i,j) is sum of paths to those two cells, as shown in steps 7-10.

Why does the loop start from i=1 and j=1 instead of 0?

Because the first row and first column are already initialized with 1s, so we start filling from the second row and column to avoid overwriting those base cases, as seen in the code and execution_table steps.

Visual Quiz - 3 Questions

Look at the execution_table at step 10, what is the value computed for dp[2][2]?

A 6

B 5

C 3

D 4

💡 Hint

Check the 'Value Computed' column at step 10 in execution_table.

At which step does the condition to compute dp[1][2] happen?

A Step 7

B Step 8

C Step 9

D Step 10

💡 Hint

Look for the row where 'Cell Updated' is (1,2) in execution_table.

If we did not initialize the first row with 1s, what would happen to dp[0][2]?

A It would be 2 because of addition

B It would be 1 anyway from dp[0][1]

C It would remain 0 or undefined, causing wrong path counts

D It would cause a runtime error

💡 Hint

Refer to variable_tracker and initialization steps 4-6 in execution_table.

Concept Snapshot

Unique Paths DP:
- Use a 2D dp array of size m x n
- Initialize first row and column to 1
- For each cell dp[i][j], dp[i][j] = dp[i-1][j] + dp[i][j-1]
- Result is dp[m-1][n-1]
- Counts unique ways moving only right/down

Full Transcript

This visualization traces the dynamic programming solution for counting unique paths in a grid. We start by initializing the first row and first column cells to 1, because from the start cell, you can only move right or down along these edges, so there is exactly one path to each. Then, for each other cell, we compute the number of unique paths by adding the paths from the cell above and the cell to the left. This builds up the dp table step-by-step until we reach the bottom-right cell, which holds the total number of unique paths. The execution table shows each step updating the dp table, and the variable tracker follows key dp cells' values. Common confusions include why the first row and column are initialized to 1, why we add the two neighboring cells, and why loops start from 1. The quiz questions test understanding of these steps and values.

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