Minimum Path Sum in Grid in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the first cell of the dp array with the grid's starting value.

DSA C

dp[0][0] = [1];

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Agrid[0][0]

B0

C1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing dp[0][0] with 0 or 1 instead of grid[0][0].

Forgetting to initialize dp[0][0].

The minimum path sum starts at the top-left cell, so dp[0][0] must be initialized with grid[0][0].

2fill in blank

medium

Complete the code to fill the first row of the dp array by accumulating sums from the left.

DSA C

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

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 dp[0][j+1] instead of dp[0][j-1].

The minimum path sum for the first row is the sum of all previous cells plus the current grid value.

3fill in blank

hard

Fix the error in the code to correctly compute the minimum path sum for each cell.

DSA C

dp[i][j] = grid[i][j] + (dp[i-1][j] < dp[i][j-1] ? dp[i-1][j] : [1]);

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Adp[i][j]

Bdp[i-1][j-1]

Cdp[i][j-1]

Dgrid[i][j]

Attempts:

3 left

💡 Hint

Common Mistakes

Using dp[i-1][j-1] which is diagonal and not allowed.

Using dp[i][j] which is the current cell and causes wrong calculation.

We choose the smaller of the top or left cell's dp values to add to the current grid cell.

4fill in blank

hard

Fill both blanks to complete the nested loops that fill the dp array starting from row 1 and column 1.

DSA C

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

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

A1

B0

C2

Dm

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loops from 0 which overwrites initial values.

Starting from 2 which skips some cells.

We start filling dp from row 1 and column 1 because row 0 and column 0 are already initialized.

5fill in blank

hard

Fill all three blanks to complete the function that returns the minimum path sum from top-left to bottom-right.

DSA C

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

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

A+

B-

Cn-1

Dm-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' instead of '+' in dp initialization.

Returning dp[m-1][m-1] instead of dp[m-1][n-1].

We add values when filling first column and row, and return dp[m-1][n-1] as the minimum path sum.

Practice

1. What does the "Minimum Path Sum in Grid" problem ask you to find?
Given a grid of numbers, find the path from top-left to bottom-right with the smallest sum. Which is true?

easy

A. Find the path with the smallest sum moving only right or down.

B. Find the path with the largest sum moving in any direction.

C. Find the path with the smallest sum moving diagonally only.

D. Find the path with the smallest sum moving left or up.

Solution

1. Step 1: Understand the movement constraints

   The problem allows moving only right or down from each cell in the grid.

2. Step 2: Identify the goal

   The goal is to find the path from the top-left corner to the bottom-right corner with the minimum sum of all numbers on the path.

3. Final Answer:

   Find the path with the smallest sum moving only right or down. -> Option A

4. Quick Check:

   Minimum path sum = smallest sum with right/down moves [OK]

Hint: Remember: only right or down moves allowed [OK]

Common Mistakes:

• Thinking diagonal moves are allowed
• Looking for maximum sum instead of minimum
• Moving left or up which is not allowed

2. Which of the following C code snippets correctly initializes a 2D array named grid with 3 rows and 3 columns filled with zeros?

easy

A. int grid[3][3] = {1};

B. int grid[3][3] = {{1}};

C. int grid[3][3]; memset(grid, 0, sizeof(grid));

D. int grid[3][3] = { {0,0,0}, {0,0,0}, {1,1,1} };

Solution

1. Step 1: Understand array initialization in C

   Initializing a 2D array with zeros can be done by setting all elements to zero using memset after declaration.

2. Step 2: Check each option

   int grid[3][3]; memset(grid, 0, sizeof(grid)); declares the array and then uses memset to set all bytes to zero, correctly initializing all elements to zero.

3. Final Answer:

   int grid[3][3]; memset(grid, 0, sizeof(grid)); -> Option C

4. Quick Check:

   memset sets all elements to zero [OK]

Hint: Use memset for zeroing arrays in C [OK]

Common Mistakes:

• Using {1} only initializes first element
• Confusing single braces with double braces
• Not including string.h for memset

3. What is the output of the following C code that calculates the minimum path sum in a 2x2 grid?

int grid[2][2] = {{1, 3}, {1, 5}};
int dp[2][2];
dp[0][0] = grid[0][0];
for(int i=1; i<2; i++) dp[0][i] = dp[0][i-1] + grid[0][i];
for(int i=1; i<2; i++) dp[i][0] = dp[i-1][0] + grid[i][0];
for(int i=1; i<2; i++) {
  for(int j=1; j<2; j++) {
    dp[i][j] = (dp[i-1][j] < dp[i][j-1] ? dp[i-1][j] : dp[i][j-1]) + grid[i][j];
  }
}
printf("%d\n", dp[1][1]);

medium

A. 10

B. 8

C. 9

D. 7

Solution

1. Step 1: Calculate first row and column sums

   dp[0][0] = 1
   dp[0][1] = 1 + 3 = 4
   dp[1][0] = 1 + 1 = 2

2. Step 2: Calculate dp[1][1]

   Minimum of dp[0][1] = 4 and dp[1][0] = 2 is 2
   dp[1][1] = 2 + 5 = 7

3. Final Answer:

   7 -> Option D

4. Quick Check:

   Minimum path sum = 7 [OK]

Hint: Fill dp row and column first, then inner cells [OK]

Common Mistakes:

• Adding max instead of min
• Incorrect dp initialization
• Off-by-one errors in loops

4. The following code attempts to compute the minimum path sum in a 3x3 grid but has a bug. What is the bug?

int grid[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
int dp[3][3];
dp[0][0] = grid[0][0];
for(int i=1; i<3; i++) dp[0][i] = dp[0][i-1] + grid[0][i];
for(int i=1; i<3; i++) dp[i][0] = dp[i-1][0] + grid[i][0];
for(int i=1; i<3; i++) {
  for(int j=1; j<3; j++) {
    dp[i][j] = dp[i-1][j] + dp[i][j-1] + grid[i][j];
  }
}

medium

A. The dp update adds both dp[i-1][j] and dp[i][j-1] instead of taking the minimum.

B. The loops start from 1 instead of 0.

C. The dp array is not initialized to zero.

D. The grid values are not accessed correctly.

Solution

1. Step 1: Analyze dp update formula

   The code adds dp[i-1][j] and dp[i][j-1] together plus grid[i][j], which is incorrect.

2. Step 2: Correct dp update

   It should take the minimum of dp[i-1][j] and dp[i][j-1], then add grid[i][j].

3. Final Answer:

   The dp update adds both dp[i-1][j] and dp[i][j-1] instead of taking the minimum. -> Option A

4. Quick Check:

   Use min(dp[i-1][j], dp[i][j-1]) + grid[i][j] [OK]

Hint: Use min, not sum, of top and left dp cells [OK]

Common Mistakes:

• Adding dp values instead of min
• Wrong loop ranges
• Not initializing dp[0][0]

5. Given a 3x3 grid:

int grid[3][3] = {
  {1, 3, 1},
  {1, 5, 1},
  {4, 2, 1}
};

Write the minimum path sum from top-left to bottom-right moving only right or down. What is the minimum sum?

hard

A. 8

B. 7

C. 9

D. 10

Solution

1. Step 1: Calculate dp first row and column

   dp[0][0] = 1
   dp[0][1] = 1+3=4
   dp[0][2] = 4+1=5
   dp[1][0] = 1+1=2
   dp[2][0] = 2+4=6

2. Step 2: Calculate inner dp cells

   dp[1][1] = min(dp[0][1], dp[1][0]) + 5 = min(4,2)+5=7
   dp[1][2] = min(dp[0][2], dp[1][1]) + 1 = min(5,7)+1=6
   dp[2][1] = min(dp[1][1], dp[2][0]) + 2 = min(7,6)+2=8
   dp[2][2] = min(dp[1][2], dp[2][1]) + 1 = min(6,8)+1=7

3. Final Answer:

   7 -> Option B

4. Quick Check:

   Minimum path sum = 7 [OK]

Hint: Fill dp row-wise and column-wise, then inner cells [OK]

Common Mistakes:

• Choosing wrong min between top and left
• Incorrect dp initialization
• Misreading grid values