Why Minimum Path Sum in Grid in DSA Python?

The Big Idea

What if you could instantly find the cheapest way through a maze of streets 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 obstacles. You want to find the cheapest path to reach your destination.

The Problem

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

The Solution

Using the Minimum Path Sum method, you can quickly calculate the cheapest path by adding costs step-by-step, remembering the best way to each block. This avoids checking all paths and finds the answer fast.

Before vs After

✗ Before

def find_min_path(grid):
    # Try all paths manually (very slow)
    pass  # no real code here because it's too complex

✓ After

def min_path_sum(grid):
    for i in range(len(grid)):
        for j in range(len(grid[0])):
            # update grid with min path sums
            pass

What It Enables

This method lets you find the cheapest route through any grid quickly and reliably, even when the grid is large.

Real Life Example

Delivery drivers use similar calculations to find the quickest or cheapest route through city streets to save time and fuel.

Key Takeaways

Manual path checking is slow and error-prone.

Minimum Path Sum uses step-by-step addition to find the cheapest path.

This approach works well for grids of any size and complexity.

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 best describes the path allowed?

easy

A. Only moves left or up at each step

B. Moves in any direction including diagonals

C. Only moves right or down at each step

D. Moves randomly without restrictions

Solution

1. Step 1: Understand the problem constraints

   The problem states you can only move right or down from each cell in the grid.

2. Step 2: Identify the path type

   This means the path is restricted to moves that go either to the right or downward, no other directions allowed.

3. Final Answer:

   Only moves right or down at each step -> Option C

4. Quick Check:

   Allowed moves = right or down [OK]

Hint: Remember: path moves only right or down [OK]

Common Mistakes:

• Thinking diagonal moves are allowed
• Assuming moves can go left or up
• Ignoring movement restrictions

2. Which of the following Python code snippets correctly initializes a 2D list named dp with zeros for a grid of size m rows and n columns?

easy

A. dp = [[0]*m for _ in range(n)]

B. dp = [[0]*n for _ in range(m)]

C. dp = [[0]*n]*m

D. dp = [0]*m*n

Solution

1. Step 1: Understand 2D list initialization

   We want a list with m rows and n columns, so outer list length is m, inner list length is n.

2. Step 2: Check each option

   dp = [[0]*n for _ in range(m)] uses list comprehension to create m separate lists each of length n filled with 0, which is correct. dp = [[0]*m for _ in range(n)] swaps m and n. dp = [[0]*n]*m creates shallow copies causing shared inner lists. dp = [0]*m*n creates a flat list, not 2D.

3. Final Answer:

   dp = [[0]*n for _ in range(m)] -> Option B

4. Quick Check:

   Correct 2D zero list = [[0]*n for _ in range(m)] [OK]

Hint: Use list comprehension for independent inner lists [OK]

Common Mistakes:

• Using [[0]*n]*m causing shared references
• Swapping m and n dimensions
• Creating a flat list instead of 2D

3. What is the output of the following code?

grid = [[1,3,1],[1,5,1],[4,2,1]]
dp = [[0]*3 for _ in range(3)]
dp[0][0] = grid[0][0]
for i in range(1,3):
    dp[0][i] = dp[0][i-1] + grid[0][i]
for j in range(1,3):
    dp[j][0] = dp[j-1][0] + grid[j][0]
for i in range(1,3):
    for j in range(1,3):
        dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]
print(dp[2][2])

medium

A. 9

B. 8

C. 10

D. 7

Solution

1. Step 1: Fill first row and column of dp

   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 dp values for inner 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 D

4. Quick Check:

   Minimum path sum = 7 [OK]

Hint: Fill dp row and column first, then inner cells [OK]

Common Mistakes:

• Mixing up indices i and j
• Adding instead of taking minimum
• Incorrect initialization of dp

4. Find the error in this code snippet for minimum path sum:

def minPathSum(grid):
    m, n = len(grid), len(grid[0])
    dp = [[0]*n]*m
    dp[0][0] = grid[0][0]
    for i in range(1, m):
        dp[i][0] = dp[i-1][0] + grid[i][0]
    for j in range(1, n):
        dp[0][j] = dp[0][j-1] + grid[0][j]
    for i in range(1, m):
        for j in range(1, n):
            dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]
    return dp[m-1][n-1]

medium

A. dp initialization uses shallow copy causing wrong updates

B. Loops start from 0 instead of 1

C. Return statement returns wrong cell

D. grid dimensions are swapped

Solution

1. Step 1: Analyze dp initialization

   dp = [[0]*n]*m creates m references to the same inner list, so updating one row updates all rows.

2. Step 2: Consequence of shallow copy

   When dp[i][0] or dp[0][j] is updated, all rows change, causing incorrect path sums.

3. Final Answer:

   dp initialization uses shallow copy causing wrong updates -> Option A

4. Quick Check:

   Use list comprehension for dp initialization [OK]

Hint: Avoid [[0]*n]*m; use list comprehension instead [OK]

Common Mistakes:

• Using shallow copy for 2D lists
• Incorrect loop ranges
• Returning wrong dp cell

5. Given a grid:

grid = [[2,1,3],[6,5,4],[7,8,9]]

Write a function to find the minimum path sum from top-left to bottom-right moving only right or down. What is the minimum path sum?

hard

A. 19

B. 21

C. 20

D. 18

Solution

1. Step 1: Initialize dp and fill first row and column

   dp[0][0]=2; first row: dp[0][1]=2+1=3, dp[0][2]=3+3=6; first column: dp[1][0]=2+6=8, dp[2][0]=8+7=15

2. Step 2: Fill remaining dp cells

   dp[1][1]=min(dp[0][1],dp[1][0])+5=min(3,8)+5=8; dp[1][2]=min(dp[0][2],dp[1][1])+4=min(6,8)+4=10; dp[2][1]=min(dp[1][1],dp[2][0])+8=min(8,15)+8=16; dp[2][2]=min(dp[1][2],dp[2][1])+9=min(10,16)+9=19

3. Final Answer:

   19 -> Option A

4. Quick Check:

   Minimum path sum = 19 [OK]

Hint: Calculate dp row and column first, then inner cells [OK]

Common Mistakes:

• Incorrect dp initialization
• Mixing up indices in loops
• Adding max instead of min