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26. Path With Minimum Effort.cpp
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26. Path With Minimum Effort.cpp
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/*
Path With Minimum Effort
========================
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 difference in 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:
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:
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:
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] <= 106
Hint #1
Consider the grid as a graph, where adjacent cells have an edge with cost of the difference between the cells.
Hint #2
If you are given threshold k, check if it is possible to go from (0, 0) to (n-1, m-1) using only edges of ≤ k cost.
Hint #3
Binary search the k value.
*/
class Solution
{
public:
int minimumEffortPath(vector<vector<int>> &a)
{
int n = a.size();
int m = a[0].size();
int v[n + 1][m + 1];
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
v[i][j] = INT_MAX;
}
}
v[0][0] = 0;
queue<pair<int, int>> q;
q.push({0, 0});
int d[5] = {0, -1, 0, 1, 0};
while (!q.empty())
{
pair<int, int> u = q.front();
q.pop();
for (int i = 0; i < 4; i++)
{
int x = u.first + d[i];
int y = u.second + d[i + 1];
if (x < 0 || y < 0 || y >= m || x >= n)
continue;
if (v[x][y] <= v[u.first][u.second])
continue;
int p = max(v[u.first][u.second], abs(a[x][y] - a[u.first][u.second]));
v[x][y] = min(p, v[x][y]);
q.push({x, y});
}
}
return v[n - 1][m - 1];
}
};