Problem Description
You are given an n x n integer matrix grid where each value grid[i][j] represents the elevation at that point (i, j).
The rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t. You can swim infinite distances in zero time. Of course, you must stay within the boundaries of the grid during your swim.
Return the least time until you can reach the bottom right square (n - 1, n - 1) if you start at the top left square (0, 0).
Solution
We can solve this by making using of Dijkstra’s Algorithm and mofifying it a bit. We are not looking at the shortest distance to reach destination Node, rather, we want to reduce the maximum level present in a path which connects source to the destination.
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class Solution {
public int swimInWater(int[][] grid) {
int n = grid.length;
int m = grid[0].length;
//to mark if a node has been visited before
int[][] visited = new int[n][m];
//Min Heap: based on level of the nodes
PriorityQueue<Node> pq = new PriorityQueue<>( (p1,p2) -> p1.level - p2.level );
//init: add the source node
pq.add(new Node(grid[0][0], 0, 0));
//while minheap is not empty
while(!pq.isEmpty()){
//get the Node with minimum level
Node current = pq.poll();
//mark that node as visited
visited[current.row][current.column] = 1;
//if that Node is the destination, we can return its level
if(current.row == n-1 && current.column == m-1) return current.level;
//Otherwise, look at the next 4 directions
int[] x = {0, 0, 1, -1};
int[] y = {1, -1, 0, 0};
//visit neighbours in the 4 directions
for(int k=0; k<4; k++){
//next row
int i = current.row + x[k];
//next column
int j = current.column + y[k];
//check if they are within bounds and not already visited
if(i>=0 && j>=0 && i<n && j<m && visited[i][j] == 0){
//the level of this Node will be max between its level and the maximum level which we have calculated to reach this Node, which is stored in current.level
int level = Math.max(current.level, grid[i][j]);
//add it to minHeap
pq.add(new Node(level, i, j));
}
}
}
return -1;
}
}
class Node{
int level;
int row;
int column;
Node(int l, int r, int c){
level = l;
row = r;
column = c;
}
}