Minimum Path Sum in Grid in DSA C++ - Deep Dive

Overview - Minimum Path Sum in Grid

What is it?

Minimum Path Sum in Grid is a problem where you find the smallest total cost to move from the top-left corner to the bottom-right corner of a grid. Each cell in the grid has a number representing the cost to step on it. You can only move right or down at each step. The goal is to add up the costs along a path that has the lowest sum.

Why it matters

This problem helps us understand how to find the best route when moving through a grid with costs, which is common in maps, games, and resource planning. Without this concept, we might waste time or resources by choosing inefficient paths. It teaches how to break down a big problem into smaller steps and find the best solution quickly.

Where it fits

Before this, you should know basic arrays and loops. After this, you can learn dynamic programming and pathfinding algorithms like Dijkstra's or A*. This problem is a stepping stone to more complex optimization problems.

Mental Model

Core Idea

Find the cheapest path by adding the smallest cost from the start to each cell, building up to the destination.

Think of it like...

Imagine walking through a city grid where each block has a toll fee. You want to pay the least total toll from your home to your friend's house by only walking east or south.

Grid example:
┌───┬───┬───┐
│ 1 │ 3 │ 1 │
├───┼───┼───┤
│ 1 │ 5 │ 1 │
├───┼───┼───┤
│ 4 │ 2 │ 1 │
└───┴───┴───┘

Path sums build up like:
┌────┬────┬────┐
│ 1  │ 4  │ 5  │
├────┼────┼────┤
│ 2  │ 7  │ 6  │
├────┼────┼────┤
│ 6  │ 8  │ 7  │
└────┴────┴────┘

Build-Up - 7 Steps

1

Foundation

Understanding the Grid and Moves

Concept: Learn what the grid looks like and the allowed moves.

You have a 2D grid with numbers. You start at the top-left cell (0,0). You can only move right (to the next column) or down (to the next row). Each move adds the cost of the cell you move onto.

Result

You know how to move through the grid and what costs you add.

Knowing allowed moves limits the problem and helps plan the path step-by-step.

2

Foundation

Calculating Path Cost by Simple Addition

Concept: Add costs along a chosen path to find total cost.

Pick a path from start to end by moving right or down. Add the numbers in each cell you step on. This sum is the path cost. Different paths have different sums.

Result

You can calculate the cost of any path you choose.

Understanding how path cost accumulates is key to comparing paths.

3

Intermediate

Brute Force Path Enumeration

🤔Before reading on: Do you think listing all paths is efficient for large grids? Commit to yes or no.

Concept: Try all possible paths to find the minimum sum.

List every path from top-left to bottom-right by moving right or down. Calculate each path's sum. Pick the smallest sum. This works but grows very fast as grid size increases.

Result

You find the minimum path sum but with slow performance on big grids.

Knowing brute force works but is slow motivates smarter methods.

4

Intermediate

Dynamic Programming Table Setup

🤔Before reading on: Will storing minimum sums for each cell help avoid repeated work? Commit to yes or no.

Concept: Use a table to store minimum path sums to each cell to avoid recalculations.

Create a 2D table same size as grid. Each cell stores the minimum sum to reach it. Start with the top-left cell equal to its cost. Fill the first row and column by adding costs cumulatively since only one path exists there.

Result

You have a table with partial minimum sums ready to build on.

Storing results prevents repeating calculations and speeds up the solution.

5

Intermediate

Filling DP Table with Recurrence

🤔Before reading on: For each cell, is the minimum sum the smaller of the top or left neighbor plus current cost? Commit to yes or no.

Concept: Calculate each cell's minimum sum using neighbors' sums plus current cell cost.

For each cell not in first row or column, minimum sum = min(top cell sum, left cell sum) + current cell cost. Fill the table row by row or column by column until bottom-right is reached.

Result

The bottom-right cell contains the minimum path sum for the whole grid.

Using neighbors' minimum sums builds the solution step-by-step efficiently.

6

Advanced

Optimizing Space Complexity

🤔Before reading on: Can we reduce memory by storing only one row or column at a time? Commit to yes or no.

Concept: Use a single array to store minimum sums for the current row to save memory.

Instead of a 2D table, keep one 1D array representing minimum sums for the current row. Update it as you move through rows, using previous values and current cell costs. This reduces space from O(m*n) to O(n).

Result

You get the same minimum path sum using less memory.

Reducing space makes the solution practical for very large grids.

7

Expert

Handling Obstacles and Variable Moves

🤔Before reading on: Can the same DP approach adapt if some cells are blocked or moves differ? Commit to yes or no.

Concept: Extend the method to grids with blocked cells or different allowed moves by adjusting conditions.

If some cells are blocked (cannot step on), mark their sums as infinite or unreachable. Modify recurrence to skip blocked cells. For different moves (e.g., diagonal), include those neighbors in the min calculation. This generalizes the approach.

Result

You can find minimum path sums even in complex grids with obstacles or extra moves.

Understanding the flexibility of DP lets you solve many grid path problems beyond the basic case.

Under the Hood

The solution uses dynamic programming, which breaks the problem into smaller subproblems. Each cell's minimum path sum depends only on the minimum sums of its immediate neighbors (top and left). By storing these results, the algorithm avoids recalculating paths multiple times, making it efficient. Internally, it fills a table or array with cumulative minimum sums, building up to the final answer.

Why designed this way?

Dynamic programming was chosen because brute force checking all paths is too slow. The problem has overlapping subproblems and optimal substructure, perfect for DP. Alternatives like recursion without memoization cause repeated work. The design balances time and space efficiency by using tabulation and optional space optimization.

┌───────────────┐
│ Start (0,0)   │
│ cost = grid[0][0] │
└──────┬────────┘
       │
       ▼
┌───────────────┐
│ DP Table Cell │
│ minSum = min( │
│ top, left) +  │
│ current cost  │
└──────┬────────┘
       │
       ▼
┌───────────────┐
│ Bottom-right  │
│ cell holds   │
│ minimum path │
│ sum          │
└───────────────┘

Myth Busters - 4 Common Misconceptions

Quick: Is the minimum path always unique? Commit to yes or no.

Common Belief: There is only one unique minimum path sum and path.

Reality: Multiple different paths can have the same minimum sum.

Why it matters: Assuming uniqueness can cause errors when reconstructing paths or counting solutions.

Quick: Can you move diagonally in the basic problem? Commit to yes or no.

Common Belief: You can move in any direction including diagonally.

Reality: Only right and down moves are allowed in the basic problem.

Why it matters: Allowing other moves changes the problem and solution approach.

Quick: Does the minimum path sum always equal the sum of the smallest numbers in the grid? Commit to yes or no.

Common Belief: The minimum path sum is just the sum of the smallest numbers in the grid.

Reality: The path must be continuous moving right or down; you cannot pick arbitrary smallest numbers.

Why it matters: Ignoring path constraints leads to incorrect answers.

Quick: Does using recursion without memoization solve the problem efficiently? Commit to yes or no.

Common Belief: Simple recursion without storing results is efficient enough.

Reality: It causes repeated calculations and exponential time complexity.

Why it matters: Leads to slow solutions that don't scale to large grids.

Expert Zone

1

The order of filling the DP table matters; row-wise or column-wise both work but must respect dependencies.

2

Space optimization is possible because each cell depends only on the current and previous row, not the entire table.

3

When reconstructing the path, backtracking from the bottom-right cell using stored decisions is needed, which is often overlooked.

When NOT to use

This approach is not suitable if moves are allowed in all directions or if costs can be negative, where algorithms like Bellman-Ford or Dijkstra's are better.

Production Patterns

Used in robotics for grid navigation, in games for pathfinding with simple movement rules, and in resource allocation problems where costs accumulate along constrained paths.

Connections

Dynamic Programming

Minimum Path Sum is a classic example of dynamic programming.

Understanding this problem helps grasp the core DP idea of building solutions from smaller subproblems.

Graph Shortest Path Algorithms

The grid can be seen as a graph where each cell is a node connected to neighbors.

Knowing graph algorithms like Dijkstra's helps extend pathfinding to more complex movement and cost scenarios.

Operations Research - Cost Minimization

Both deal with finding minimum cost paths or flows under constraints.

Recognizing this connection shows how algorithmic solutions apply to real-world optimization problems in logistics and planning.

Common Pitfalls

#1 Trying to move diagonally or in other directions not allowed.

Wrong approach:

for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]); // includes diagonal
}
}

Correct approach:

for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1]); // only top and left
}
}

Root cause:

Misunderstanding allowed moves changes the problem and solution.

#2 Not initializing the first row and first column properly.

Wrong approach:

dp[0][0] = grid[0][0];
for (int i = 1; i < m; i++) dp[i][0] = grid[i][0]; // missing cumulative addition
for (int j = 1; j < n; j++) dp[0][j] = grid[0][j]; // missing cumulative addition

Correct approach:

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

Root cause:

Forgetting that first row and column have only one path, so sums must accumulate.

#3 Using recursion without memoization causing slow performance.

Wrong approach:

int minPathSum(int i, int j) {
if (i == 0 && j == 0) return grid[0][0];
if (i < 0 || j < 0) return INT_MAX;
return grid[i][j] + min(minPathSum(i-1, j), minPathSum(i, j-1));
}

Correct approach:

Use a dp table or memo array to store results and avoid repeated calls.

Root cause:

Not storing intermediate results causes exponential calls.

Key Takeaways

Minimum Path Sum in Grid finds the cheapest route from top-left to bottom-right by moving only right or down.

Dynamic programming efficiently solves this by building a table of minimum sums for each cell using neighbors' results.

Proper initialization of the first row and column is crucial because they have only one possible path.

Space can be optimized by using a single array since each cell depends only on the previous row or column.

This problem is a foundation for understanding more complex pathfinding and optimization algorithms.

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