Day-9 Leetcode Coding Journey: Mastering Problem-Solving Techniques

LeetCode #63: Unique Paths II

Problem Statement

You are given an m x n grid where some cells contain obstacles (1) and others are free (0).
The robot starts at the top-left corner and can only move right or down.
The robot must reach the bottom-right corner of the grid.
Find the number of unique paths the robot can take while avoiding obstacles.

Constraints

• 1≤m,n≤100

• obstacleGrid[i][j] is either 0 (free cell) or 1 (obstacle).

🚀 Code Implementation

class Solution(object):
    def uniquePathsWithObstacles(self, obstacleGrid):
        """
        :type obstacleGrid: List[List[int]]
        :rtype: int
        """
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        if obstacleGrid[0][0] == 1 or obstacleGrid[m-1][n-1] == 1:
            return 0
        dp = [[0] * n for _ in range(m)]
        dp[0][0] = 1
        for i in range(1, m):
            if obstacleGrid[i][0] == 0:
                dp[i][0] = dp[i-1][0]  
        for j in range(1, n):
            if obstacleGrid[0][j] == 0:
                dp[0][j] = dp[0][j-1]  
        for i in range(1, m):
            for j in range(1, n):
                if obstacleGrid[i][j] == 0:
                    dp[i][j] = dp[i-1][j] + dp[i][j-1]
        return dp[m-1][n-1]

LeetCode #64: Minimum Path Sum

Problem Statement

Given an m x n grid filled with non-negative numbers, find a path from the top-left to the bottom-right corner that minimizes the sum of all numbers along the path.
You can only move right or down at any point in time.

Constraints

• 1≤m,n≤100

• 0≤grid[i][j]≤100

🚀 Code Implementation

class Solution(object):
    def minPathSum(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        m, n = len(grid), len(grid[0])
        for j in range(1, n):
            grid[0][j] += grid[0][j - 1]
        for i in range(1, m):
            grid[i][0] += grid[i - 1][0]
        for i in range(1, m):
            for j in range(1, n):
                grid[i][j] += min(grid[i - 1][j], grid[i][j - 1])        
        return grid[m - 1][n - 1]