3342. Find Minimum Time to Reach Last Room II.cpp

[shafaq16]

Approach:

This problem uses Dijkstra’s algorithm on a grid to find the minimum time to reach the bottom-right cell from the top-left.
Each cell has a moveTime constraint — you can’t enter it before that time.
We use a min-heap (priority queue) to always expand the cell that can be reached earliest.
At each step:
Pop the cell with the smallest current time.
Try moving in all 4 directions (up, down, left, right).
For each neighbor:
Compute wait time = max(moveTime[i_][j_] - currTime, 0) (if we must wait before entering).
Compute move cost = 1 or 2 depending on cell parity (i_ + j_) % 2.
Update arrival time and push to the queue if it’s smaller than the previous best.

Intuition:

It’s like a weighted shortest path problem on a grid.
You can’t move into a cell before its “unlock time” (moveTime[i][j]), so you might need to wait.
Using Dijkstra ensures we always choose the earliest possible time to reach each cell while considering both waiting and moving costs efficiently.

Solution in Code(C++)
class Solution {
public:
vector<vector>directions{{1,0},{-1,0},{0,1},{0,-1}};
typedef pair<int,pair<int,int>>P;
int minTimeToReach(vector<vector>& moveTime) {
int m=moveTime.size(),n=moveTime[0].size();

    priority_queue<P,vector<P>,greater<P>>pq;
    vector<vector<int>>result(m,vector<int>(n,INT_MAX));

    result[0][0]=0;
    pq.push({0,{0,0}});

    while(!pq.empty()){
        int currTime=pq.top().first;
        auto cell=pq.top().second;

        int i=cell.first;
        int j=cell.second;

        pq.pop();
        if(i==m-1 && j==n-1) return currTime;

        for(auto &dir:directions){
            int i_=i+dir[0];
            int j_=j+dir[1];

            if(i_>=0 && i_<m && j_>=0 && j_<n){
                int moveCost=(i_+j_)%2==0?2:1;
                int wait=max(moveTime[i_][j_]-currTime,0);
                int arrTime=currTime+wait+moveCost;

                if(result[i_][j_]>arrTime){
                    result[i_][j_]=arrTime;
                    pq.push({arrTime,{i_,j_}});
                }
            }
        }
    }
    return -1;
}

};