Why Minimum Path Sum in Grid in DSA C?

The Big Idea

What if you could instantly find the cheapest way through a maze of city blocks without trying every path?

The Scenario

Imagine you are walking through a city grid from the top-left corner to the bottom-right corner. Each block has a cost to cross, like traffic or tolls. You want to find the cheapest path to reach your destination.

The Problem

Trying every possible path manually is slow and confusing because the number of paths grows very fast. You might miss the cheapest route or waste time checking paths that are too expensive.

The Solution

The Minimum Path Sum method uses a smart way to remember the cheapest cost to reach each block. It builds up the answer step by step, so you don't repeat work and quickly find the cheapest path.

Before vs After

✗ Before

int minPathSum(int grid[][100], int rows, int cols) {
    // Try all paths recursively - very slow
    // No memory of past results
}

✓ After

int minPathSum(int grid[][100], int rows, int cols) {
    int dp[rows][cols];
    // Fill dp with minimum sums step by step
    // Use dp to avoid repeated work
}

What It Enables

This method lets you quickly find the cheapest route in a grid, saving time and effort even for large maps.

Real Life Example

Delivery drivers use this to find the cheapest route through city blocks with different traffic costs, saving fuel and time.

Key Takeaways

Manual checking of all paths is too slow and error-prone.

Dynamic programming remembers past results to speed up finding the minimum path sum.

This approach helps in real-world route planning and cost optimization.

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