16. Path With Minimum Effort

The problem can be found at the following link: Question Link

My Approach

1. Start by defining a class named Solution with a public member function minimumEffortPath, which takes a 2D vector heights as input and returns an integer.

2. Get the dimensions of the heights grid: m represents the number of rows, and n represents the number of columns.

3. Check if the grid has only one cell (m == 1 and n == 1), return 0 as there is no effort required to move within a single-cell grid.

4. Create a 2D vector dist of size m x n to store the minimum effort required to reach each cell from the starting cell. Initialize all values in dist to INT_MAX except for the starting cell, which is set to 0.

5. Create a priority_queue named pq to store cells as pairs of (distance, cell), where distance represents the minimum effort to reach that cell, and the cell is represented as (i, j).

6. Define two vectors, di and dj, to represent the four possible directions to move: up, left, down, and right. These vectors will be used to explore neighboring cells.

7. Start a while loop that continues as long as the priority queue pq is not empty.

8. Inside the loop, pop the top element from pq, which represents the cell with the smallest effort required.

9. Extract the row i and column j of the current cell from the popped element.

10. Iterate through the four possible directions (up, left, down, right) using a for loop.

11. For each direction, calculate the new row ni and new column nj by adding the corresponding values from di and dj to the current row i and column j.

12. Check if the new cell (ni, nj) is within the boundaries of the grid (i.e., ni >= 0, nj >= 0, ni < m, and nj < n).

13. Calculate the new distance newDist required to reach the new cell. This distance is the maximum of the current distance dist[i][j] and the absolute difference in heights between the current cell and the new cell abs(heights[i][j] - heights[ni][nj]).

14. If the new distance newDist is smaller than the current distance dist[ni][nj], update dist[ni][nj] with newDist and push the new cell (ni, nj) into the priority queue pq with the updated distance.

15. Repeat steps 8-14 until all cells have been processed, ensuring that the priority queue always contains cells with the minimum effort.

16. Finally, return the minimum effort required to reach the bottom-right cell, which is stored in dist[m - 1][n - 1].

This algorithm uses Dijkstra's algorithm with a priority queue to find the minimum effort path from the top-left corner to the bottom-right corner of the grid while considering the differences in heights between adjacent cells.

Time and Auxiliary Space Complexity

• Time Complexity: O(m*n*log(m*n))

• Auxiliary Space Complexity: O(m*n)

Code (C++)

class Solution {
public:
    int minimumEffortPath(std::vector<std::vector<int>>& heights) {
        int m = heights.size();
        int n = heights[0].size();
        if (m == 1 && n == 1) {
            return 0;
        }

        vector<vector<int>> dist(m, vector<int>(n, INT_MAX));
        dist[0][0] = 0;


        priority_queue<pair<int, int>, vector<pair<int, int>>,greater<pair<int, int>>> pq;
        pq.push({0, 0});
  
        vector<int> di = {-1, 0, 1, 0};
        vector<int> dj = {0, -1, 0, 1};

        while (!pq.empty()) {
            auto it = pq.top();
            pq.pop();
            int i = it.first;
            int j = it.second;

            for (int k = 0; k < 4; k++) {
                int ni = i + di[k];
                int nj = j + dj[k];

                if (ni >= 0 && nj >= 0 && ni < m && nj < n) {

                    // Important logic of the entire code
                    int newDist = max(dist[i][j], abs(heights[i][j] - heights[ni][nj]));

                    // If we found a shorter path, update distance and push to priority queue
                    if (newDist < dist[ni][nj]) {
                        dist[ni][nj] = newDist;
                        pq.push({ni, nj});
                    }
                }
            }
        }
        return dist[m - 1][n - 1];
    }
};

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