I like to read this solution as a small machine: keep the useful information, throw away the noise. For 1594. Maximum Non Negative Product in a Matrix, the solution in this repository is mainly a DFS + memoization solution.
Guide
What?
The code is easier to read if we treat it as a controlled search through possible states. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: DFS + memoization, dynamic programming.
The notes already sitting in the source point us in the right direction:
- dfs with memorization
Guide
When?
Use this approach when the hard part is not syntax, but deciding what must stay true after every update. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are dfs, dfs_back, maxProductPath.
Guide
Why?
The win comes from making each line carry responsibility: store the useful state, discard the rest, keep moving.
- The two-dimensional vector is the memory of the solution: grid state, DP state, or adjacency shape.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Initialize the memory or helper structure.
- Process candidates in the order the invariant expects.
- Update the answer only when the current state is valid.
- Return the value that represents the fully processed input.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(n) to O(n log n), depending on the dominant loop or data structure operation
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
solution.cpp
01//dfs with memorization
02//Runtime: 24 ms, faster than 16.04% of C++ online submissions for Maximum Non Negative Product in a Matrix.
03//Memory Usage: 15.1 MB, less than 5.04% of C++ online submissions for Maximum Non Negative Product in a Matrix.
04class Solution {
05public:
06 int m, n;
07 long long maxprod;
08 vector<vector<vector<long long>>> memo;
09
10 void dfs(vector<vector<int>>& grid, int r, int c, long long prod){
11 prod *= grid[r][c];
12 // cout << r << ", " << c << endl;
13 if(r == m-1 && c == n-1){
14 // cout << prod << endl;
15 maxprod = max(maxprod, prod);
16 }else{
17 if(r+1 < m){
18 dfs(grid, r+1, c, prod);
19 }
20
21 if(c+1 < n){
22 dfs(grid, r, c+1, prod);
23 }
24 }
25 };
26
27 vector<long long> dfs_back(vector<vector<int>>& grid, int endr, int endc){
28 // cout << endr << ", " << endc << endl; //", " << pos << endl;
29 if(endr == 0 && endc == 0){
30 return memo[0][0] = {grid[0][0], grid[0][0]};
31 }else if(memo[endr][endc][0] != std::numeric_limits<long long>::lowest()){
32 return memo[endr][endc];
33 }else{
34 long long maxres = std::numeric_limits<long long>::lowest();
35 long long minres = std::numeric_limits<long long>::max();
36 // res = grid[endr][endc];
37
38 // bool newpos = pos;
39 // if(grid[endr][endc] < 0) newpos = !newpos;
40
41 bool pos = grid[endr][endc] > 0;
42 if(endr-1 >= 0){
43 vector<long long> ret = dfs_back(grid, endr-1, endc);
44 maxres = max(maxres, pos ? ret[1] * grid[endr][endc] : ret[0] * grid[endr][endc]);
45 minres = min(minres, pos ? ret[0] * grid[endr][endc] : ret[1] * grid[endr][endc]);
46 }
47 if(endc-1 >= 0){
48 vector<long long> ret = dfs_back(grid, endr, endc-1);
49 maxres = max(maxres, pos ? ret[1] * grid[endr][endc] : ret[0] * grid[endr][endc]);
50 minres = min(minres, pos ? ret[0] * grid[endr][endc] : ret[1] * grid[endr][endc]);
51 }
52
53 return memo[endr][endc] = {minres, maxres};
54 }
55 };
56
57 int maxProductPath(vector<vector<int>>& grid) {
58 m = grid.size();
59 n = grid[0].size();
60 maxprod = std::numeric_limits<long long>::lowest();
61 memo = vector<vector<vector<long long>>>(m, vector<vector<long long>>(n, vector<long long>(2, std::numeric_limits<long long>::lowest())));
62
63 int MOD = 1e9 + 7;
64
65 long long prod = 1LL;
66 // dfs(grid, 0, 0, prod);
67 // if(maxprod < 0) return -1;
68 // return maxprod % MOD;
69
70 vector<long long> res = dfs_back(grid, m-1, n-1);
71
72// for(int i = 0; i < m; ++i){
73// for(int j = 0; j < n; ++j){
74// cout << memo[i][j][0] << " ";
75// }
76// cout << endl;
77// }
78
79// for(int i = 0; i < m; ++i){
80// for(int j = 0; j < n; ++j){
81// cout << memo[i][j][1] << " ";
82// }
83// cout << endl;
84// }
85
86 if(res[1] < 0) return -1;
87 return res[1] % MOD;
88 }
89};
90
91//bottom-up DP
92//https://leetcode.com/problems/maximum-non-negative-product-in-a-matrix/discuss/855082/C%2B%2B-Dynamic-Programming-With-Comments
93//Runtime: 8 ms, faster than 92.88% of C++ online submissions for Maximum Non Negative Product in a Matrix.
94//Memory Usage: 10.7 MB, less than 15.98% of C++ online submissions for Maximum Non Negative Product in a Matrix.
95class Solution {
96public:
97 int maxProductPath(vector<vector<int>>& grid) {
98 int m = grid.size();
99 int n = grid[0].size();
100 int MOD = 1e9 + 7;
101
102 vector<vector<long long>> mx(m, vector<long long>(n, 0LL));
103 vector<vector<long long>> mn(m, vector<long long>(n, 0LL));
104
105 //base case
106 mx[0][0] = mn[0][0] = grid[0][0];
107
108 //in the leftmost column, we only have one choice
109 for(int i = 1; i < m; ++i){
110 mx[i][0] = mn[i][0] = mx[i-1][0] * grid[i][0];
111 }
112
113 //in the top row, we only have one choice
114 for(int j = 1; j < n; ++j){
115 mx[0][j] = mn[0][j] = mn[0][j-1] * grid[0][j];
116 }
117
118 for(int i = 1; i < m; ++i){
119 for(int j = 1; j < n; ++j){
120 if(grid[i][j] > 0){
121 //the product could be negative, so don't MOD here
122 mx[i][j] = max(mx[i-1][j], mx[i][j-1]) * grid[i][j];
123 mn[i][j] = min(mn[i-1][j], mn[i][j-1]) * grid[i][j];
124 }else{
125 mx[i][j] = min(mn[i-1][j], mn[i][j-1]) * grid[i][j];
126 mn[i][j] = max(mx[i-1][j], mx[i][j-1]) * grid[i][j];
127 }
128 }
129 }
130
131 return mx[m-1][n-1] < 0LL ? -1 : mx[m-1][n-1] % MOD;
132 }
133};Cost
Complexity
Time
O(n) to O(n log n), depending on the dominant loop or data structure operation
Dominated by the main traversal, recursion, or data-structure operations in the code.
Space
O(n) in the usual case for auxiliary containers or recursion
Auxiliary state plus the answer structure where the problem requires one.