2023-08-12 Daily Challenge

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In: DailyChallenge

Today I have done leetcode's August LeetCoding Challenge with cpp.

August LeetCoding Challenge 12

Description

Unique Paths II

You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.

Return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The testcases are generated so that the answer will be less than or equal to 2 * 109.

 

Example 1:

Input: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
Output: 2
Explanation: There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

Example 2:

Input: obstacleGrid = [[0,1],[0,0]]
Output: 1

 

Constraints:

  • m == obstacleGrid.length
  • n == obstacleGrid[i].length
  • 1 <= m, n <= 100
  • obstacleGrid[i][j] is 0 or 1.

Solution

class Solution {
public:
  int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
    int m = obstacleGrid.size(), n = obstacleGrid[0].size();
    if(obstacleGrid[0][0] || obstacleGrid[m-1][n-1]) return 0;
    vector<int> dp(m*n);
    dp[0] = 1;
    for(int i = 1; i < m*n; ++i) {
      if(obstacleGrid[i/n][i%n] == 1) continue;
      if(i >= n) dp[i] += dp[i-n];
      if(i % n != 0) dp[i] += dp[i-1];
    }
    return dp.back();
  }
};

// Accepted
// 41/41 cases passed (5 ms)
// Your runtime beats 28.03 % of cpp submissions
// Your memory usage beats 61.67 % of cpp submissions (7.7 MB)
Algorithm