1631. Path With Minimum Effort

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1631. Path With Minimum Effort

Description

Difficulty: Medium

Related Topics: Array, Binary Search, Depth-First Search, Breadth-First Search, Union Find, Heap (Priority Queue), Matrix

You are a hiker preparing for an upcoming hike. You are given heights, a 2D array of size rows x columns, where heights[row][col] represents the height of cell (row, col). You are situated in the top-left cell, (0, 0), and you hope to travel to the bottom-right cell, (rows-1, columns-1) (i.e., 0-indexed). You can move up, down, left, or right, and you wish to find a route that requires the minimum effort.

A route’s effort is the maximum absolute differencein heights between two consecutive cells of the route.

Return the minimum effort required to travel from the top-left cell to the bottom-right cell.

Example 1:

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Input: heights = [[1,2,2],[3,8,2],[5,3,5]]
Output: 2
Explanation: The route of [1,3,5,3,5] has a maximum absolute difference of 2 in consecutive cells.
This is better than the route of [1,2,2,2,5], where the maximum absolute difference is 3.

Example 2:

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Input: heights = [[1,2,3],[3,8,4],[5,3,5]]
Output: 1
Explanation: The route of [1,2,3,4,5] has a maximum absolute difference of 1 in consecutive cells, which is better than route [1,3,5,3,5].

Example 3:

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Input: heights = [[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]]
Output: 0
Explanation: This route does not require any effort.

Constraints:

  • rows == heights.length
  • columns == heights[i].length
  • 1 <= rows, columns <= 100
  • 1 <= heights[i][j] <= 10<sup>6</sup>

Hints/Notes

  • Dijkstra algorithm: BFS with dp table

Solution

Language: C++

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class Solution {
public:
    int m, n;

    int minimumEffortPath(vector<vector<int>>& heights) {
        m = heights.size(); n = heights[0].size();

        vector<vector<int>> efforts(m, vector<int>(n, INT_MAX));

        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;

        pq.push({0, 0, 0});
        while (!pq.empty()) {
            vector<int> point = pq.top();
            pq.pop();
            int i = point[1];
            int j = point[2];
            int e = point[0];
            if (i == m - 1 && j == n - 1) {
                return e;
            }
            if (e > efforts[i][j]) {
                continue;
            }
            vector<pair<int, int>>points = adj(i, j);
            for (auto point : points) {
                int nextI = point.first;
                int nextJ = point.second;
                int effort = efforts[nextI][nextJ];
                int newEffort = max(abs(heights[nextI][nextJ] - heights[i][j]), e);
                if (effort > newEffort) {
                    efforts[nextI][nextJ] = newEffort;
                    pq.push({newEffort, nextI, nextJ});
                }
            }
        }
        return -1;
    }

    vector<pair<int, int>> adj(int i, int j) {
        vector<pair<int, int>> res;
        if (i - 1 >= 0) {
            res.push_back({i - 1, j});
        }
        if (i + 1 < m) {
            res.push_back({i + 1, j});
        }
        if (j - 1 >= 0) {
            res.push_back({i, j - 1});
        }
        if (j + 1 < n) {
            res.push_back({i, j + 1});
        }
        return res;
    }
};