Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
Note: You can only move either down or right at any point in time.
Example:
Input: [ [1,3,1], [1,5,1], [4,2,1] ] Output: 7 Explanation: Because the path 1→3→1→1→1 minimizes the sum.
Your algorithm should run in O(n * m) time and use O(n * m) extra space.
The core challenge of this problem is to find the path from the top-left corner to the bottom-right corner of a grid that results in the minimum sum of the numbers along the path. The constraints are that you can only move either down or right at any point in time.
This problem is significant in various applications such as robotics (finding the shortest path in a grid), game development (pathfinding algorithms), and operations research (optimizing routes).
Potential pitfalls include misunderstanding the movement constraints and not considering all possible paths.
To solve this problem, we can use dynamic programming. The idea is to build a 2D array dp where dp[i][j] represents the minimum path sum to reach cell (i, j).
1. **Naive Solution**: A naive approach would involve exploring all possible paths using recursion, which would be highly inefficient due to the exponential number of paths.
2. **Optimized Solution**: Using dynamic programming, we can build the solution iteratively. We initialize the top-left cell with its value and then fill in the rest of the dp table by considering the minimum path sum to reach each cell from either the top or the left.
1. Initialize a 2D array dp with the same dimensions as the input grid.
2. Set dp[0][0] to grid[0][0].
3. Fill in the first row and first column of dp since they can only be reached from one direction.
4. For each cell (i, j) in the grid, set dp[i][j] to the value of the grid at that cell plus the minimum of the values from the cell above or the cell to the left.
5. The value at dp[m-1][n-1] will be the minimum path sum.
def minPathSum(grid):
# Get the dimensions of the grid
m, n = len(grid), len(grid[0])
# Create a 2D dp array with the same dimensions as grid
dp = [[0] * n for _ in range(m)]
# Initialize the top-left cell
dp[0][0] = grid[0][0]
# Fill in the first row
for j in range(1, n):
dp[0][j] = dp[0][j-1] + grid[0][j]
# Fill in the first column
for i in range(1, m):
dp[i][0] = dp[i-1][0] + grid[i][0]
# Fill in the rest of the dp array
for i in range(1, m):
for j in range(1, n):
dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1])
# The bottom-right cell contains the minimum path sum
return dp[m-1][n-1]
# Example usage
grid = [
[1, 3, 1],
[1, 5, 1],
[4, 2, 1]
]
print(minPathSum(grid)) # Output: 7
The time complexity of this approach is O(n * m) because we iterate through each cell of the grid once. The space complexity is also O(n * m) due to the additional dp array.
1. **Single Cell Grid**: The grid has only one cell. The output should be the value of that cell.
2. **Single Row or Column**: The grid has only one row or one column. The path is straightforward, and the sum is the sum of all cells in that row or column.
3. **Large Values**: The grid contains large values. Ensure the algorithm handles large sums without overflow.
To test the solution comprehensively, consider the following test cases:
[[5]][[1, 2, 3]][[1], [2], [3]][[1, 3, 1], [1, 5, 1], [4, 2, 1]][[1000, 1000], [1000, 1000]]1. **Break Down the Problem**: Understand the constraints and break down the problem into smaller subproblems.
2. **Use Dynamic Programming**: For optimization problems, dynamic programming can often reduce the time complexity significantly.
3. **Practice**: Solve similar problems to get a better grasp of dynamic programming techniques.
In this blog post, we discussed the problem of finding the minimum path sum in a grid. We explored a dynamic programming approach to solve the problem efficiently. Understanding and solving such problems is crucial for developing strong problem-solving skills in computer science.