1970. Last Day Where You Can Still Cross LeetCode Solution
In this guide, you will get 1970. Last Day Where You Can Still Cross LeetCode Solution with the best time and space complexity. The solution to Last Day Where You Can Still Cross problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Last Day Where You Can Still Cross
There is a 1-based binary matrix where 0 represents land and 1 represents water. You are given integers row and col representing the number of rows and columns in the matrix, respectively.
Initially on day 0, the entire matrix is land. However, each day a new cell becomes flooded with water. You are given a 1-based 2D array cells, where cells[i] = [ri, ci] represents that on the ith day, the cell on the rith row and cith column (1-based coordinates) will be covered with water (i.e., changed to 1).
You want to find the last day that it is possible to walk from the top to the bottom by only walking on land cells. You can start from any cell in the top row and end at any cell in the bottom row. You can only travel in the four cardinal directions (left, right, up, and down).
Return the last day where it is possible to walk from the top to the bottom by only walking on land cells.
Input: row = 2, col = 2, cells = [[1,1],[2,1],[1,2],[2,2]]
Output: 2
Explanation: The above image depicts how the matrix changes each day starting from day 0.
The last day where it is possible to cross from top to bottom is on day 2.
Example 2:
Input: row = 2, col = 2, cells = [[1,1],[1,2],[2,1],[2,2]]
Output: 1
Explanation: The above image depicts how the matrix changes each day starting from day 0.
The last day where it is possible to cross from top to bottom is on day 1.
Example 3:
Input: row = 3, col = 3, cells = [[1,2],[2,1],[3,3],[2,2],[1,1],[1,3],[2,3],[3,2],[3,1]]
Output: 3
Explanation: The above image depicts how the matrix changes each day starting from day 0.
The last day where it is possible to cross from top to bottom is on day 3.
Constraints:
2 <= row, col <= 2 * 104
4 <= row * col <= 2 * 104
cells.length == row * col
1 <= ri <= row
1 <= ci <= col
All the values of cells are unique.
Complexity Analysis
Time Complexity: O(\texttt{row} \cdot \texttt{col} \cdot \log (\texttt{row} \cdot \texttt{col}))
Space Complexity: O(\texttt{row} \cdot \texttt{col})
1970. Last Day Where You Can Still Cross LeetCode Solution in C++
class Solution {
public:
int latestDayToCross(int row, int col, vector<vector<int>>& cells) {
int ans = 0;
int l = 1;
int r = cells.size() - 1;
while (l <= r) {
const int m = (l + r) / 2;
if (canWalk(m, row, col, cells)) {
ans = m;
l = m + 1;
} else {
r = m - 1;
}
}
return ans;
}
private:
static constexpr int dirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
bool canWalk(int day, int row, int col, const vector<vector<int>>& cells) {
vector<vector<int>> matrix(row, vector<int>(col));
for (int i = 0; i < day; ++i) {
const int x = cells[i][0] - 1;
const int y = cells[i][1] - 1;
matrix[x][y] = 1;
}
queue<pair<int, int>> q;
for (int j = 0; j < col; ++j)
if (matrix[0][j] == 0) {
q.emplace(0, j);
matrix[0][j] = 1;
}
while (!q.empty()) {
const auto [i, j] = q.front();
q.pop();
for (const auto& [dx, dy] : dirs) {
const int x = i + dx;
const int y = j + dy;
if (x < 0 || x == row || y < 0 || y == col)
continue;
if (matrix[x][y] == 1)
continue;
if (x == row - 1)
return true;
q.emplace(x, y);
matrix[x][y] = 1;
}
}
return false;
}
}; /* code provided by PROGIEZ */
1970. Last Day Where You Can Still Cross LeetCode Solution in Java
class Solution {
public int latestDayToCross(int row, int col, int[][] cells) {
int ans = 0;
int l = 1;
int r = cells.length - 1;
while (l <= r) {
final int m = (l + r) / 2;
if (canWalk(m, row, col, cells)) {
ans = m;
l = m + 1;
} else {
r = m - 1;
}
}
return ans;
}
private static final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
private boolean canWalk(int day, int row, int col, int[][] cells) {
int[][] matrix = new int[row][col];
for (int i = 0; i < day; ++i) {
final int x = cells[i][0] - 1;
final int y = cells[i][1] - 1;
matrix[x][y] = 1;
}
Queue<Pair<Integer, Integer>> q = new ArrayDeque<>();
for (int j = 0; j < col; ++j)
if (matrix[0][j] == 0) {
q.offer(new Pair<>(0, j));
matrix[0][j] = 1;
}
while (!q.isEmpty()) {
final int i = q.peek().getKey();
final int j = q.poll().getValue();
for (int[] dir : dirs) {
final int x = i + dir[0];
final int y = j + dir[1];
if (x < 0 || x == row || y < 0 || y == col)
continue;
if (matrix[x][y] == 1)
continue;
if (x == row - 1)
return true;
q.offer(new Pair<>(x, y));
matrix[x][y] = 1;
}
}
return false;
}
} // code provided by PROGIEZ
1970. Last Day Where You Can Still Cross LeetCode Solution in Python
class Solution:
def latestDayToCross(self, row: int, col: int, cells: list[list[int]]) -> int:
dirs = ((0, 1), (1, 0), (0, -1), (-1, 0))
def canWalk(day: int) -> bool:
matrix = [[0] * col for _ in range(row)]
for i in range(day):
x, y = cells[i]
matrix[x - 1][y - 1] = 1
q = collections.deque()
for j in range(col):
if matrix[0][j] == 0:
q.append((0, j))
matrix[0][j] = 1
while q:
i, j = q.popleft()
for dx, dy in dirs:
x = i + dx
y = j + dy
if x < 0 or x == row or y < 0 or y == col:
continue
if matrix[x][y] == 1:
continue
if x == row - 1:
return True
q.append((x, y))
matrix[x][y] = 1
return False
ans = 0
l = 1
r = len(cells) - 1
while l <= r:
m = (l + r) // 2
if canWalk(m):
ans = m
l = m + 1
else:
r = m - 1
return ans # code by PROGIEZ
