Maximum Product Path in Matrix - Google Top Interview Questions
Problem Statement :
You are given a two-dimensional list of integers matrix.
You are currently at the top left corner and want to move to the bottom right corner.
In each move, you can move down or right.
Return the maximum product of the cells visited by going to the bottom right cell.
If the result is negative, return -1. Otherwise, mod the result by 10 ** 9 + 7.
Constraints
1 ≤ n, m ≤ 20 where n and m are the number of rows and columns in matrix
-2 ≤ matrix[r][c] ≤ 2
Example 1
Input
matrix = [
[2, 1, -2],
[-1, -1, -2],
[1, 1, 1]
]
Output
8
Explanation
We can take the following path: [2, 1, -2, -2, 1].
Solution :
Solution in C++ :
typedef long long ll;
int solve(vector<vector<int>>& matrix) {
int n = matrix.size(), m = matrix[0].size();
vector<vector<pair<ll, ll>>> f(n, vector<pair<ll, ll>>(m));
f[0][0] = {matrix[0][0], matrix[0][0]};
for (int i = 0; i < n; ++i)
for (int j = 0; j < m; ++j) {
ll lo = LLONG_MAX, hi = LLONG_MIN;
if (i > 0) {
lo = min(lo,
min(f[i - 1][j].first * matrix[i][j], f[i - 1][j].second * matrix[i][j]));
hi = max(hi,
max(f[i - 1][j].first * matrix[i][j], f[i - 1][j].second * matrix[i][j]));
}
if (j > 0) {
lo = min(lo,
min(f[i][j - 1].first * matrix[i][j], f[i][j - 1].second * matrix[i][j]));
hi = max(hi,
max(f[i][j - 1].first * matrix[i][j], f[i][j - 1].second * matrix[i][j]));
}
if (i || j) f[i][j] = {lo, hi};
}
return f[n - 1][m - 1].second >= 0 ? f[n - 1][m - 1].second % 1000000007 : -1;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] mat) {
int n = mat.length, m = mat[0].length;
long dp[][][] = new long[n][m][2];
dp[0][0][0] = dp[0][0][1] = mat[0][0];
// max,min
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (i == 0 && j == 0)
continue;
dp[i][j][0] = Long.MIN_VALUE;
dp[i][j][1] = Long.MAX_VALUE;
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (i == 0 && j == 0)
continue;
if (j - 1 >= 0) {
int cur = mat[i][j];
dp[i][j][0] = Math.max(dp[i][j][0], cur * dp[i][j - 1][0]);
dp[i][j][0] = Math.max(dp[i][j][0], cur * dp[i][j - 1][1]);
dp[i][j][1] = Math.min(dp[i][j][1], cur * dp[i][j - 1][0]);
dp[i][j][1] = Math.min(dp[i][j][1], cur * dp[i][j - 1][1]);
}
if (i - 1 >= 0) {
int cur = mat[i][j];
dp[i][j][0] = Math.max(dp[i][j][0], cur * dp[i - 1][j][0]);
dp[i][j][0] = Math.max(dp[i][j][0], cur * dp[i - 1][j][1]);
dp[i][j][1] = Math.min(dp[i][j][1], cur * dp[i - 1][j][0]);
dp[i][j][1] = Math.min(dp[i][j][1], cur * dp[i - 1][j][1]);
}
}
}
return (dp[n - 1][m - 1][0] < 0) ? -1 : (int) (dp[n - 1][m - 1][0] % 1000000007);
}
}
Solution in Python :
class Solution:
def mod(self, x):
if x < 0:
return -1 * (abs(x) % int(1e9 + 7))
return x % int(1e9 + 7)
def solve(self, mat):
@cache
def f(r, c):
m, n = len(mat), len(mat[0])
if r == m - 1 and c == n - 1:
return mat[r][c], mat[r][c]
mymin, mymax, cur = 1000, -1000, mat[r][c]
if r + 1 < m:
nmax, nmin = f(r + 1, c)
mymin = min(mymin, cur * nmax, cur * nmin)
mymax = max(mymax, cur * nmax, cur * nmin)
if c + 1 < n:
nmax, nmin = f(r, c + 1)
mymin = min(mymin, cur * nmax, cur * nmin)
mymax = max(mymax, cur * nmax, cur * nmin)
return mymax, mymin
hi, lo = f(0, 0)
return max(self.mod(hi), self.mod(lo), -1)
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