Longest Matrix Path Length - Google Top Interview Questions
Problem Statement :
You are given a two dimensional integer matrix where 0 is an empty cell and 1 is a wall.
You can start at any empty cell on row 0 and want to end up on any empty cell on row n - 1.
Given that you can move left, right, or down, return the longest such path where you visit each cell at most once. If there is no viable path, return 0.
Constraints
1 ≤ n * m ≤ 200,000 where n and m are the number of rows and columns in matrix.
Example 1
Input
matrix = [
[0, 0, 0, 0],
[1, 0, 0, 0],
[0, 0, 0, 0]
]
Output
10
Explanation
We can move (0, 0), (0, 1), (0, 2), (0, 3), (1, 3), (1, 2), (1, 1), (2, 1), (2, 2), (2, 3).
Solution :
Solution in C++ :
bool isValid(int i, int j, int n, int m) {
if (i >= 0 && i < n && j >= 0 && j < m) return true;
return false;
}
// 0->left,1->right,2->down
// left = i,j-1
// down = i+1,j
// right = i,j+1
int rec(int f, int i, int j, vector<vector<int>>& mat, vector<vector<vector<int>>>& dp, int n,
int m) {
if (i == n - 1) {
if (f == 0 && j == 0) return 0;
if (f == 1 && j == m - 1) return 0;
}
int& ans = dp[f][i][j];
if (ans != -2) return ans;
ans = 0;
if (f == 0) {
// I am in left direction ,hence I can only go left and down
if (isValid(i, j - 1, n, m) && mat[i][j - 1] == 0)
ans = 1 + rec(0, i, j - 1, mat, dp, n, m);
if (isValid(i + 1, j, n, m) && mat[i + 1][j] == 0)
ans = max(ans, 1 + rec(2, i + 1, j, mat, dp, n, m));
} else if (f == 1) {
// I am in right direction , hence I can only go further right and down
if (isValid(i, j + 1, n, m) && mat[i][j + 1] == 0)
ans = 1 + rec(1, i, j + 1, mat, dp, n, m);
if (isValid(i + 1, j, n, m) && mat[i + 1][j] == 0)
ans = max(ans, rec(2, i + 1, j, mat, dp, n, m) + 1);
} else {
// now I am in downward direction hence I can go in all three directions
// cout<<"hel";
if (isValid(i, j + 1, n, m) && mat[i][j + 1] == 0)
ans = 1 + rec(1, i, j + 1, mat, dp, n, m);
if (isValid(i + 1, j, n, m) && mat[i + 1][j] == 0) {
ans = max(ans, 1 + rec(2, i + 1, j, mat, dp, n, m));
// cout<<"hey";
}
if (isValid(i, j - 1, n, m) && mat[i][j - 1] == 0)
ans = max(ans, 1 + rec(0, i, j - 1, mat, dp, n, m));
}
// cout<<ans;
if (ans == 0 && i != n - 1) return ans = -1;
return ans;
}
int solve(vector<vector<int>>& mat) {
int n = mat.size();
if (!n) return 0;
int m = mat[0].size();
// cout<<n<<endl;
vector<vector<vector<int>>> dp(3, vector<vector<int>>(n, vector<int>(m, -2)));
int ans = 0;
for (int j = 0; j < m; j++) {
if (mat[0][j] == 1) continue;
// cout<<"hello
";
// cout<<rec(2,0,j,mat,dp,n,m);
ans = max(ans, 1 + rec(2, 0, j, mat, dp, n, m));
}
return ans;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] matrix) {
int N = matrix.length;
int M = matrix[0].length;
int[][] dp = new int[N + 1][M];
for (int i = 0; i < N; i++) Arrays.fill(dp[i], Integer.MIN_VALUE);
for (int i = N - 1; i >= 0; i--) {
for (int j = 0; j < M; j++) {
if (matrix[i][j] == 1)
continue;
for (int k = j; k >= 0; k--) {
if (matrix[i][k] == 1)
break;
int x = 1 + Math.abs(j - k);
dp[i][j] = Math.max(dp[i][j], x + dp[i + 1][k]);
}
for (int k = j; k < M; k++) {
if (matrix[i][k] == 1)
break;
int x = 1 + Math.abs(j - k);
dp[i][j] = Math.max(dp[i][j], x + dp[i + 1][k]);
}
}
}
int ans = 0;
for (int j = 0; j < M; j++) ans = Math.max(ans, dp[0][j]);
return ans;
}
}
Solution in Python :
class Solution:
def solve(self, matrix):
rows = len(matrix)
cols = len(matrix[0])
ninf = -1000000000
@cache
def dfs(x, y, dy):
if x == rows:
return 0
if matrix[x][y] == 1:
return ninf
best = ninf
if dy == 0:
best = max(best, dfs(x, y, 1))
best = max(best, dfs(x, y, -1))
elif 0 <= y + dy < cols:
best = max(best, dfs(x, y + dy, dy) + 1)
best = max(best, dfs(x + 1, y, 0) + 1)
return best
best = 0
for y in range(cols):
best = max(best, dfs(0, y, 0))
return best
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