Min Cost Path
Detailed guide and Python implementation for the 'Min Cost Path' problem.
1. Concept Overview
The 'Min Cost Path' problem is a key challenge in the Dynamic Programming section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Min Cost Path.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Min Cost Path carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Write a function min_cost_path(cost, m, n) that finds the minimum cost to reach cell (m, n) from cell (0, 0) in a 2D cost matrix cost. You can only move down, right, and diagonally down-right from a cell.
- •1 <= len(cost), len(cost[0]) <= 100
- •0 <= cost[i][j] <= 1000
- •0 <= m < len(cost)
- •0 <= n < len(cost[0])
Examples
min_cost_path([[1, 2, 3], [4, 8, 2], [1, 5, 3]], 2, 2)
8
The path with minimum cost is (0,0) -> (0,1) -> (1,2) -> (2,2) with total cost 1 + 2 + 2 + 3 = 8.
min_cost_path([[1, 2, 3], [4, 8, 2], [1, 5, 3]], 1, 1)
9
The path with minimum cost is (0,0) -> (1,1) with total cost 1 + 8 = 9.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def min_cost_opt(cost, m, n):
tc = [[0] * (n + 1) for _ in range(m + 1)]
tc[0][0] = cost[0][0]
for i in range(1, m + 1): tc[i][0] = tc[i-1][0] + cost[i][0]
for j in range(1, n + 1): tc[0][j] = tc[0][j-1] + cost[0][j]
for i in range(1, m + 1):
for j in range(1, n + 1):
tc[i][j] = min(tc[i-1][j-1], tc[i-1][j], tc[i][j-1]) + cost[i][j]
return tc[m][n]Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def min_cost_brute(cost, m, n):
if n < 0 or m < 0: return float('inf')
if m == 0 and n == 0: return cost[m][n]
return cost[m][n] + min(min_cost_brute(cost, m-1, n-1), min_cost_brute(cost, m-1, n), min_cost_brute(cost, m, n-1))Algorithm Pattern Checklist
When dealing with Dynamic Programming data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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