3341. Find Minimum Time to Reach Last Room I.cpp

[shafaq16]

Approach:

This solution also uses Dijkstra’s algorithm to find the minimum time to reach the destination cell.
Each cell (i, j) can only be entered after its given moveTime[i][j].
We use a priority queue to always expand the cell that can be reached at the earliest time.
Steps:
Start from (0, 0) at time 0.
Use a min-heap to process cells in order of increasing arrival time.
For each direction (up, down, left, right):
You can move to (nrow, ncol) only after moveTime[nrow][ncol].
The earliest time you can arrive there is max(curr, moveTime[nrow][ncol]) + 1.
If this time is smaller than the previously recorded distance, update it and push it into the heap.

Intuition:

Think of the grid as a map with time-locked cells — you can’t step on a cell before its allowed time.
Using Dijkstra ensures we always choose the minimum total time path considering both waiting (if the cell isn’t ready) and moving (which always costs +1).
It’s like finding the fastest route through a grid where some roads open later than others.

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

    priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<>> pq;
    pq.push({0, {0, 0}});

    vector<vector<int>> dis(m, vector<int>(n, INT_MAX));
    dis[0][0] = 0;

    int dir[5] = {-1, 0, 1, 0, -1};

    while (!pq.empty()) {
        auto [curr, pos] = pq.top(); pq.pop();
        int row = pos.first, col = pos.second;

        if (row == m - 1 && col == n - 1)
            return curr;

        for (int i = 0; i < 4; i++) {
            int nrow = row + dir[i], ncol = col + dir[i + 1];
            if (nrow >= 0 && ncol >= 0 && nrow < m && ncol < n) {
                int nextTime = max(curr, moveTime[nrow][ncol]) + 1;
                if (nextTime < dis[nrow][ncol]) {
                    dis[nrow][ncol] = nextTime;
                    pq.push({nextTime, {nrow, ncol}});
                }
            }
        }
    }

    return -1;
}

};

[SjxSubham]

U and others who contributed already are only allowed to solve only 2 issue...
Let's give chance to others as well...
#119 #118 #120