Minimum Path Sum in Grid in DSA Go - 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 sum of numbers from the top-left corner to the bottom-right corner of a grid. You can only move either down or right at any point in time. The grid contains non-negative numbers, and the goal is to find the path that adds up to the smallest sum. This helps in understanding how to optimize paths in a grid-like structure.

Why it matters

This problem helps solve real-world tasks like finding the cheapest route on a map or minimizing costs in a grid-based system. Without this concept, we might waste time or resources by choosing inefficient paths. It teaches how to break down problems into smaller steps and build up solutions efficiently.

Where it fits

Before this, you should know basic arrays and loops. After this, you can learn dynamic programming and more complex pathfinding algorithms like Dijkstra's or A*.

Mental Model

Core Idea

Find the smallest sum path by choosing the best step at each grid cell, moving only right or down.

Think of it like...

Imagine walking through a city grid where each block has a toll cost. You want to pay the least total toll from your home (top-left) to your friend's house (bottom-right), moving only east or south.

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

Path example: 1 -> 3 -> 1 -> 1 -> 1
Sum: 7 (minimum path sum)

Build-Up - 7 Steps

1

Foundation

Understanding the Grid and Moves

Concept: Introduce the grid structure and allowed moves (right and down).

A grid is a 2D array of numbers. You start at the top-left cell (0,0). From any cell, you can move only to the cell on the right or the cell below. The goal is to reach the bottom-right cell (last row, last column).

Result

You know how to navigate the grid step-by-step using only right or down moves.

Understanding allowed moves limits the problem space and simplifies path choices.

2

Foundation

Calculating Path Sums by Hand

Concept: Manually add numbers along a path to find its total sum.

Pick a path from top-left to bottom-right by moving right or down. Add all numbers on the path. For example, in a 3x3 grid: 1 3 1 1 5 1 4 2 1 Path: (0,0)->(0,1)->(0,2)->(1,2)->(2,2) Sum: 1+3+1+1+1=7

Result

You can find the sum of any path by adding numbers along it.

Knowing how to calculate path sums manually helps understand what the algorithm must optimize.

3

Intermediate

Brute Force Path Enumeration

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

Concept: Try all possible paths and pick the minimum sum path.

There are many paths from top-left to bottom-right. For each path, calculate the sum and remember the smallest. This works but is slow because paths grow exponentially with grid size.

Result

You get the correct minimum sum but with very slow performance on big grids.

Understanding brute force shows why we need smarter methods to handle bigger grids efficiently.

4

Intermediate

Dynamic Programming Table Setup

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

Concept: Use a table to store minimum sums to each cell, building up from the start.

Create a 2D array dp of the same size as the grid. dp[i][j] will hold the minimum sum to reach cell (i,j). Initialize dp[0][0] with grid[0][0]. For first row and column, sums accumulate from left or top respectively.

Result

You have a structure that remembers best sums for each cell, avoiding repeated calculations.

Knowing that subproblems overlap allows us to store and reuse results, making the solution efficient.

5

Intermediate

Filling DP Table with Recurrence

🤔Before reading on: Should dp[i][j] be the minimum of dp[i-1][j] and dp[i][j-1] plus grid[i][j]? Commit to yes or no.

Concept: Calculate dp[i][j] by choosing the smaller sum from top or left neighbor plus current cell value.

For each cell (i,j), dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1]). This means the best path to (i,j) comes from the best path to either the cell above or to the left.

Result

The dp table fills with minimum sums for all cells, culminating in dp[last_row][last_col] as the answer.

Understanding the recurrence relation is key to building the solution bottom-up.

6

Advanced

Optimizing Space Usage

🤔Before reading on: Can we reduce dp space from 2D to 1D array? Commit to yes or no.

Concept: Use a single array to store minimum sums for the current row, updating it as we move down.

Since dp[i][j] depends only on dp[i-1][j] and dp[i][j-1], we can keep one array representing the current row's dp values. Update it in place while iterating through rows.

Result

Memory usage reduces from O(m*n) to O(n), where n is the number of columns.

Knowing dependencies lets us optimize memory without losing correctness.

7

Expert

Handling Obstacles and Variations

🤔Before reading on: Can the same dp approach handle grids with blocked cells? Commit to yes or no.

Concept: Modify dp to skip or assign infinite cost to blocked cells, adapting the algorithm to more complex grids.

If a cell is blocked, set dp value to a large number or skip updating it. This way, paths through blocked cells are ignored. The rest of the dp logic remains the same.

Result

The algorithm can find minimum path sums even with obstacles, or report no path if blocked.

Understanding how to adapt the core dp method to variations makes it robust for real-world problems.

Under the Hood

The algorithm uses dynamic programming to build solutions from smaller subproblems. Each cell's minimum path sum depends only on the sums of its immediate top and left neighbors. By storing these intermediate results, it avoids recalculating paths multiple times, reducing exponential complexity to polynomial time.

Why designed this way?

Dynamic programming was designed to solve problems with overlapping subproblems and optimal substructure. This problem fits perfectly because the best path to a cell depends only on best paths to adjacent cells. Alternatives like brute force were too slow, so DP offers an efficient, systematic approach.

Grid indices and dp values:

  (0,0)   (0,1)   (0,2)
+-------+-------+-------+
| dp[0][0] | dp[0][1] | dp[0][2] |
+-------+-------+-------+
| dp[1][0] | dp[1][1] | dp[1][2] |
+-------+-------+-------+
| dp[2][0] | dp[2][1] | dp[2][2] |
+-------+-------+-------+

Each dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1])
Edges handle boundaries by using only one neighbor.

Myth Busters - 4 Common Misconceptions

Quick: Does the minimum path always go through the smallest numbers in the grid? Commit yes or no.

Common Belief: The minimum path must pass through the smallest numbers in the grid.

Reality: The minimum path depends on the sum of numbers along the path, not just individual small numbers. Sometimes a path with slightly larger numbers but fewer steps or better arrangement is cheaper overall.

Why it matters: Assuming the path must include smallest numbers can lead to wrong guesses and inefficient algorithms.

Quick: Can you move diagonally in the minimum path sum problem? Commit yes or no.

Common Belief: You can move diagonally as well as right and down.

Reality: Only moves to the right or down are allowed. Diagonal moves are not permitted in this problem.

Why it matters: Allowing diagonal moves changes the problem and solution approach, leading to incorrect results if misunderstood.

Quick: Is brute force practical for large grids? Commit yes or no.

Common Belief: Brute force checking all paths is practical for any grid size.

Reality: Brute force is only feasible for very small grids because the number of paths grows exponentially with grid size.

Why it matters: Trying brute force on large grids causes huge delays and resource waste.

Quick: Does the minimum path sum always equal the sum of the first row or first column? Commit yes or no.

Common Belief: The minimum path sum is always the sum along the first row or first column.

Reality: The minimum path sum can be less than or equal to those sums because the path can combine moves right and down to find a cheaper route.

Why it matters: Assuming this limits understanding of the problem and leads to incorrect shortcuts.

Expert Zone

1

The order of filling the dp table matters; it must be row-wise or column-wise to ensure dependencies are computed first.

2

When optimizing space, careful handling of boundary conditions is required to avoid index errors.

3

In some cases, reconstructing the path from the dp table requires backtracking from the bottom-right cell using stored decisions.

When NOT to use

This approach is not suitable if diagonal or other moves are allowed, or if the grid is dynamic with changing values. For such cases, graph algorithms like Dijkstra's or A* are better alternatives.

Production Patterns

Used in robotics for grid navigation cost optimization, in game development for pathfinding on tile maps, and in logistics for minimizing transportation costs on grid-like networks.

Connections

Dynamic Programming

Minimum Path Sum is a classic example of dynamic programming applied to grid problems.

Understanding this problem deepens comprehension of dynamic programming principles like overlapping subproblems and optimal substructure.

Graph Shortest Path Algorithms

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

Recognizing the grid as a graph helps connect grid path problems to general shortest path algorithms like Dijkstra's.

Supply Chain Optimization

Both involve finding minimum cost paths through a network of steps or stages.

Knowing minimum path sum concepts helps understand cost minimization in supply chains, where each step adds cost and choices affect total expense.

Common Pitfalls

#1 Ignoring boundary conditions when filling dp table.

Wrong approach:

for i := 1; i < m; i++ {
for j := 1; j < n; j++ {
dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1])
}
}
// No initialization for first row and first column

Correct approach:

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

Root cause:

Not handling edges separately causes index errors or incorrect sums because dp values outside the grid are undefined.

#2 Using brute force for large grids.

Wrong approach:

func minPathSum(grid [][]int) int {
// Generate all paths recursively and find minimum sum
// Exponential time complexity
}

Correct approach:

func minPathSum(grid [][]int) int {
m, n := len(grid), len(grid[0])
dp := make([][]int, m)
for i := range dp {
dp[i] = make([]int, n)
}
dp[0][0] = grid[0][0]
// Fill dp with minimum sums
// Efficient O(m*n) time
}

Root cause:

Not recognizing exponential growth of paths leads to inefficient solutions.

#3 Allowing invalid moves like diagonal steps.

Wrong approach:

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

Correct approach:

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

Root cause:

Misunderstanding problem constraints causes incorrect algorithm and results.

Key Takeaways

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

Dynamic programming efficiently solves this by storing minimum sums for each cell, avoiding repeated work.

Proper handling of grid edges and initialization is crucial for correct results.

Optimizing space is possible by using a single array instead of a full 2D table.

Understanding this problem builds a foundation for more complex pathfinding and optimization algorithms.

Practice

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

easy

A. The number of paths from top-left to bottom-right moving only left or up

B. The largest sum of numbers from the top-left to bottom-right moving only diagonally

C. The smallest sum of numbers from the top-left to bottom-right moving only right or down

D. The sum of all numbers in the grid

Solution

1. Step 1: Understand the problem goal

   The problem asks to find the path from the top-left corner to the bottom-right corner of a grid that has the smallest total sum of the numbers on the path.

2. Step 2: Recognize allowed moves

   Only moves to the right or down are allowed, so the path must follow these directions.

3. Final Answer:

   The smallest sum of numbers from the top-left to bottom-right moving only right or down -> Option C

4. Quick Check:

   Minimum path sum = smallest total sum [OK]

Hint: Focus on moves right or down to find smallest sum path [OK]

Common Mistakes:

• Thinking diagonal moves are allowed
• Looking for largest sum instead of smallest
• Counting all numbers instead of path sum

2. Which of the following Go code snippets correctly initializes a 2D grid for the Minimum Path Sum problem?

easy

A. grid := [3][3]int{{1, 3, 1}, {1, 5, 1}, {4, 2, 1}}

B. grid := []int{ {1, 3, 1}, {1, 5, 1}, {4, 2, 1} }

C. grid := [][]int{1, 3, 1, 1, 5, 1, 4, 2, 1}

D. grid := [][]int{{1, 3, 1}, {1, 5, 1}, {4, 2, 1}}

Solution

1. Step 1: Check correct 2D slice initialization

   grid := [][]int{{1, 3, 1}, {1, 5, 1}, {4, 2, 1}} uses [][]int with nested slices, which is the correct way to declare a 2D slice in Go.

2. Step 2: Identify incorrect options

   grid := [3][3]int{{1, 3, 1}, {1, 5, 1}, {4, 2, 1}} uses fixed array type which is valid but less flexible; grid := [][]int{1, 3, 1, 1, 5, 1, 4, 2, 1} is a flat slice, not 2D; grid := []int{ {1, 3, 1}, {1, 5, 1}, {4, 2, 1} } uses []int with nested braces which is invalid syntax.

3. Final Answer:

   grid := [][]int{{1, 3, 1}, {1, 5, 1}, {4, 2, 1}} -> Option D

4. Quick Check:

   2D slice with nested slices = correct grid init [OK]

Hint: Use [][]int with nested braces for 2D slices [OK]

Common Mistakes:

• Using flat slices instead of 2D slices
• Incorrect array type syntax
• Missing nested braces for rows

3. What is the output of the following Go code that calculates the minimum path sum?

func minPathSum(grid [][]int) int {
    m, n := len(grid), len(grid[0])
    for i := 1; i < m; i++ {
        grid[i][0] += grid[i-1][0]
    }
    for j := 1; j < n; j++ {
        grid[0][j] += grid[0][j-1]
    }
    for i := 1; i < m; i++ {
        for j := 1; j < n; j++ {
            if grid[i-1][j] < grid[i][j-1] {
                grid[i][j] += grid[i-1][j]
            } else {
                grid[i][j] += grid[i][j-1]
            }
        }
    }
    return grid[m-1][n-1]
}

func main() {
    grid := [][]int{{1,3,1},{1,5,1},{4,2,1}}
    fmt.Println(minPathSum(grid))
}

medium

A. 7

B. 8

C. 9

D. 10

Solution

1. Step 1: Calculate first column cumulative sums

   grid[1][0] = 1+1=2, grid[2][0] = 2+4=6

2. Step 2: Calculate first row cumulative sums

   grid[0][1] = 1+3=4, grid[0][2] = 4+1=5

3. Step 3: Calculate remaining cells minimum path sums

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

4. Final Answer:

   7 -> Option A

5. Quick Check:

   Minimum path sum = 7 [OK]

Hint: Fill first row and column, then choose min of top or left [OK]

Common Mistakes:

• Not updating first row and column sums
• Adding max instead of min
• Mixing indices in loops

4. Find the error in this Go function for minimum path sum and select the fix:

func minPathSum(grid [][]int) int {
    m, n := len(grid), len(grid[0])
    for i := 1; i <= m; i++ {
        grid[i][0] += grid[i-1][0]
    }
    for j := 1; j <= n; j++ {
        grid[0][j] += grid[0][j-1]
    }
    for i := 1; i < m; i++ {
        for j := 1; j < n; j++ {
            if grid[i-1][j] < grid[i][j-1] {
                grid[i][j] += grid[i-1][j]
            } else {
                grid[i][j] += grid[i][j-1]
            }
        }
    }
    return grid[m-1][n-1]
}

medium

A. Initialize grid with zeros before loops

B. Change loops to use < instead of <= for i and j

C. Add else if condition for equal values

D. Return grid[0][0] instead of grid[m-1][n-1]

Solution

1. Step 1: Identify loop boundary error

   The loops use <= which causes index out of range error because valid indices are 0 to m-1 and 0 to n-1.

2. Step 2: Fix loop conditions

   Changing loops to use < instead of <= prevents out of range access and correctly iterates over grid indices.

3. Final Answer:

   Change loops to use < instead of <= for i and j -> Option B

4. Quick Check:

   Loop indices must be less than length [OK]

Hint: Use < length, not <= length in loops [OK]

Common Mistakes:

• Using <= causing index out of range
• Not checking array bounds
• Returning wrong grid cell

5. Given a grid with some cells having zero cost, how does the presence of zero values affect the minimum path sum calculation? Example grid:

[[1, 0, 3],
 [0, 5, 1],
 [4, 2, 0]]

Choose the correct statement.

hard

A. Zero cost cells can reduce the minimum path sum by providing cheaper paths

B. Zero cost cells are ignored in the calculation

C. Zero cost cells cause the algorithm to fail due to division by zero

D. Zero cost cells increase the minimum path sum

Solution

1. Step 1: Understand zero cost cell effect

   Cells with zero cost add no extra sum, so paths passing through them can have lower total sums.

2. Step 2: Confirm algorithm behavior

   The algorithm sums cell values; zero values reduce total sum, so zero cost cells help find cheaper paths.

3. Final Answer:

   Zero cost cells can reduce the minimum path sum by providing cheaper paths -> Option A

4. Quick Check:

   Zero cost cells lower path sum [OK]

Hint: Zero cells lower path sum by adding no cost [OK]

Common Mistakes:

• Ignoring zero cells in sum
• Assuming zero causes errors
• Thinking zero increases cost