Minimum Path Sum in Grid in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the goal of the Minimum Path Sum problem in a grid?

To find the path from the top-left corner to the bottom-right corner of a grid that has the smallest sum of all numbers along the path, moving only down or right.

beginner

Which moves are allowed when finding the minimum path sum in a grid?

Only moves to the right or moves down are allowed at each step.

intermediate

How can dynamic programming help solve the Minimum Path Sum problem?

By storing the minimum path sums to each cell, we avoid recalculating paths multiple times and build the solution step-by-step from the top-left to the bottom-right.

beginner

What is the base case when using dynamic programming for Minimum Path Sum in a grid?

The top-left cell's minimum path sum is its own value because it is the starting point.

intermediate

If the grid is [[1,3,1],[1,5,1],[4,2,1]], what is the minimum path sum?

The minimum path sum is 7, following the path 1 → 3 → 1 → 1 → 1.

In the Minimum Path Sum problem, which directions can you move?

A Right and Down

B Left and Up

C Right and Up

D Left and Down

Only moves to the right or down are allowed to find the minimum path sum.

What does dynamic programming store to solve the Minimum Path Sum efficiently?

A All possible paths

B Only the last path sum

C Maximum sums for each cell

D Minimum sums for each cell

Dynamic programming stores the minimum path sums for each cell to avoid repeated calculations.

What is the minimum path sum for a grid with only one cell containing 5?

A 5

B 1

C 0

D Cannot be determined

With only one cell, the minimum path sum is the value of that cell itself.

If the top-left cell value is 2 and the cell to its right is 3, what is the minimum path sum to the right cell?

A 2

B 5

C Cannot move right

D 3

Minimum path sum to the right cell is 2 (start) + 3 = 5.

Which of these is NOT a valid step in the Minimum Path Sum problem?

A Moving down

B Moving right

C Moving diagonally

D Starting at top-left

Diagonal moves are not allowed; only right and down moves are valid.

Explain how you would use dynamic programming to find the minimum path sum in a grid.

Think about building the solution step-by-step using previous results.

Describe the base cases and how to fill the first row and first column in the Minimum Path Sum grid.

Consider how you can only move right in the first row and only down in the first column.

Practice

1. What does the "Minimum Path Sum in Grid" problem ask you to find?

easy

A. The largest sum of numbers along any path in the grid.

B. The shortest path length from top-left to bottom-right.

C. The number of paths from top-left to bottom-right.

D. The smallest sum of numbers along a path from top-left to bottom-right, moving only right or down.

Solution

1. Step 1: Understand the problem goal

   The problem asks for the minimum sum of values collected when moving from the top-left corner to the bottom-right corner of a grid, moving only right or down.

2. Step 2: Clarify movement constraints

   Only moves to the right or down are allowed, so the path is restricted and the sum depends on chosen cells.

3. Final Answer:

   The smallest sum of numbers along a path from top-left to bottom-right, moving only right or down. -> Option D

4. Quick Check:

   Minimum path sum = smallest sum path [OK]

Hint: Focus on sum along right/down moves only [OK]

Common Mistakes:

• Thinking diagonal moves are allowed
• Looking for maximum sum instead of minimum
• Counting number of paths instead of sum

2. Which of the following C++ code snippets correctly initializes a 2D vector named grid with 3 rows and 4 columns filled with zeros?

easy

A. std::vector grid(3, std::vector<int>(4, 0));

B. std::vector<int> grid(3, std::vector<int>(4, 0));

C. std::vector grid(4, std::vector<int>(3, 0));

D. std::vector grid(3, 4, 0);

Solution

1. Step 1: Understand 2D vector initialization syntax

   To create a 2D vector with 3 rows and 4 columns filled with zeros, we use vector of vectors: outer vector size 3, inner vector size 4, each element 0.

2. Step 2: Check each option

   std::vector grid(3, std::vector<int>(4, 0)); correctly uses std::vector grid(3, std::vector<int>(4, 0)); which means 3 rows, each row is a vector of 4 zeros.

3. Final Answer:

   std::vector grid(3, std::vector<int>(4, 0)); -> Option A

4. Quick Check:

   Correct 2D vector init = std::vector grid(3, std::vector<int>(4, 0)); [OK]

Hint: Use vector of vectors with inner vector size and default value [OK]

Common Mistakes:

• Using wrong vector type for 2D
• Swapping rows and columns sizes
• Trying to pass three arguments to vector constructor

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

std::vector grid = {{1, 3}, {1, 5}};
int m = grid.size();
int n = grid[0].size();
for (int i = 1; i < m; i++) grid[i][0] += grid[i-1][0];
for (int j = 1; j < n; j++) grid[0][j] += grid[0][j-1];
for (int i = 1; i < m; i++) {
  for (int j = 1; j < n; j++) {
    grid[i][j] += std::min(grid[i-1][j], grid[i][j-1]);
  }
}
std::cout << grid[m-1][n-1] << std::endl;

medium

A. 7

B. 8

C. 9

D. 10

Solution

1. Step 1: Calculate first column cumulative sums

   grid[1][0] = grid[1][0] + grid[0][0] = 1 + 1 = 2

2. Step 2: Calculate first row cumulative sums

   grid[0][1] = grid[0][1] + grid[0][0] = 3 + 1 = 4

3. Step 3: Calculate inner cell minimum path sums

   grid[1][1] = grid[1][1] + min(grid[0][1], grid[1][0]) = 5 + min(4, 2) = 5 + 2 = 7

4. Final Answer:

   7 -> Option A

5. Quick Check:

   Minimum path sum = 7 [OK]

Hint: Update first row and column cumulatively, then inner cells [OK]

Common Mistakes:

• Not updating first row and column before inner cells
• Adding max instead of min
• Mixing indices i and j

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

std::vector grid = {{1, 2, 3}, {4, 5, 6}};
int m = grid.size();
int n = grid[0].size();
for (int i = 1; i < m; i++) grid[i][0] += grid[i-1][0];
for (int j = 1; j < n; j++) grid[0][j] += grid[0][j-1];
for (int i = 1; i < m; i++) {
  for (int j = 1; j < n; j++) {
    grid[i][j] += std::max(grid[i-1][j], grid[i][j-1]);
  }
}
std::cout << grid[m-1][n-1] << std::endl;

medium

A. Wrong loop ranges for i and j

B. Not initializing first row and column

C. Using std::max instead of std::min in inner loop

D. Incorrect grid initialization syntax

Solution

1. Step 1: Check the inner loop operation

   The code uses std::max to update grid[i][j], but the problem requires minimum path sum, so std::min should be used.

2. Step 2: Verify other parts

   First row and column are correctly initialized, loops have correct ranges, and grid initialization is valid.

3. Final Answer:

   Using std::max instead of std::min in inner loop -> Option C

4. Quick Check:

   Use min for minimum path sum [OK]

Hint: Use min, not max, when updating path sums [OK]

Common Mistakes:

• Confusing max and min functions
• Skipping first row/column initialization
• Off-by-one errors in loops

5. Given the grid below, what is the minimum path sum from top-left to bottom-right?

grid = {
  {1, 3, 1, 2},
  {1, 5, 1, 1},
  {4, 2, 1, 3}
};

You can only move right or down. Calculate the minimum sum.

hard

A. 8

B. 10

C. 9

D. 11

Solution

1. Step 1: Compute first row and first column cumulative sums

   First row: [1, 4, 5, 7] (1, 1+3=4, 4+1=5, 5+2=7) First column: [1, 2, 6] (1, 1+1=2, 2+4=6)

2. Step 2: Calculate inner cells minimum path sums

   Row 1, col 1: 5 + min(4,2) = 5 + 2 = 7 Row 1, col 2: 1 + min(5,7) = 1 + 5 = 6 Row 1, col 3: 1 + min(7,6) = 1 + 6 = 7 Row 2, col 1: 2 + min(7,6) = 2 + 6 = 8 Row 2, col 2: 1 + min(6,8) = 1 + 6 = 7 Row 2, col 3: 3 + min(7,7) = 3 + 7 = 10

3. Final Answer:

   10 -> Option B

4. Quick Check:

   Minimum path sum = 10 [OK]

Hint: Fill first row/column, then inner cells with min sums [OK]

Common Mistakes:

• Adding max instead of min
• Incorrect cumulative sums for first row/column
• Misindexing grid cells