LeetCode 1631: Path With Minimum Effort
LeetCode 1631 Solution Explanation
Explanation
To solve this problem, we can use Dijkstra's algorithm to find the minimum effort required to travel from the top-left cell to the bottom-right cell. We will create a priority queue to store the cells with their corresponding efforts, and at each step, we will explore the neighboring cells and update the effort required to reach each cell. We will keep track of the minimum effort to reach the bottom-right cell.
The algorithm steps are as follows:
- Initialize a priority queue and add the top-left cell with effort 0.
- While the priority queue is not empty:
- Pop the cell with the minimum effort from the priority queue.
- If this cell is the bottom-right cell, return its effort as the minimum effort.
- Otherwise, explore its neighbors and update the effort required to reach each neighbor.
- Return the minimum effort found.
The time complexity of this algorithm is O(rows * columns * log(rows * columns)) where rows and columns are the dimensions of the input grid. The space complexity is O(rows * columns) for storing the efforts in the priority queue.
LeetCode 1631 Solutions in Java, C++, Python
class Solution {
public int minimumEffortPath(int[][] heights) {
int rows = heights.length;
int cols = heights[0].length;
int[][] efforts = new int[rows][cols];
for (int i = 0; i < rows; i++) {
Arrays.fill(efforts[i], Integer.MAX_VALUE);
}
PriorityQueue<int[]> pq = new PriorityQueue<>(Comparator.comparingInt(a -> a[2]));
pq.offer(new int[]{0, 0, 0});
int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
while (!pq.isEmpty()) {
int[] curr = pq.poll();
int row = curr[0];
int col = curr[1];
int effort = curr[2];
if (row == rows - 1 && col == cols - 1) {
return effort;
}
for (int[] dir : dirs) {
int newRow = row + dir[0];
int newCol = col + dir[1];
if (newRow >= 0 && newRow < rows && newCol >= 0 && newCol < cols) {
int newEffort = Math.max(effort, Math.abs(heights[newRow][newCol] - heights[row][col]));
if (newEffort < efforts[newRow][newCol]) {
efforts[newRow][newCol] = newEffort;
pq.offer(new int[]{newRow, newCol, newEffort});
}
}
}
}
return -1; // Should not reach here
}
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