Why Minimum Path Sum in Grid in DSA Go?

The Big Idea

What if you could find the cheapest path through a maze of costs without trying every route?

The Scenario

Imagine you are in a city laid out as a grid. You want to walk from the top-left corner to the bottom-right corner, but each block has a cost to cross. You try to find the cheapest path by guessing and checking each route manually.

The Problem

Manually checking every possible path is slow and confusing because the number of paths grows very fast as the grid gets bigger. It's easy to make mistakes and miss cheaper routes.

The Solution

Using the Minimum Path Sum method, you can calculate the cheapest path step-by-step by remembering the best cost to reach each block. This way, you avoid repeating work and quickly find the minimum total cost.

Before vs After

✗ Before

func minPathSum(grid [][]int) int {
    // Try all paths manually - very slow
    return 0
}

✓ After

func minPathSum(grid [][]int) int {
    for i := 1; i < len(grid); i++ {
        grid[i][0] += grid[i-1][0]
    }
    for j := 1; j < len(grid[0]); j++ {
        grid[0][j] += grid[0][j-1]
    }
    for i := 1; i < len(grid); i++ {
        for j := 1; j < len(grid[0]); 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[len(grid)-1][len(grid[0])-1]
}

What It Enables

This method lets you quickly find the cheapest path through any grid, no matter how big, without checking every route.

Real Life Example

Delivery drivers use similar calculations to find the quickest or cheapest route through city streets with different traffic or toll costs.

Key Takeaways

Manual path checking is slow and error-prone.

Minimum Path Sum uses step-by-step cost updates to find the cheapest path efficiently.

This approach saves time and avoids repeated work.

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