Memoization Approach for Gold Mine Problem on GFG - c++

Question

I am trying to solve the Gold Mine problem on GFG using a memoization based approach of dynamic programming. Here is the code I have written.
int dp[50][50];
int traverse(int x,int y,int n,int m, vector<vector<int>> M){
if((x<n and x>=0) and (y<m and y>=0))
{
if(dp[x][y]!=-1)
return dp[x][y];
else{
int right=M[x][y]+traverse(x,y+1,n,m,M);
int right_up=M[x][y]+traverse(x-1,y+1,n,m,M);
int right_down=M[x][y]+traverse(x+1,y+1,n,m,M);
return dp[x][y]=max(max(right,right_up),right_down);}
}
return 0;
}
int maxGold(int n, int m, vector<vector<int>> M)
{
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
dp[i][j]=-1;
}
}
int ans=0;
for(int i=0;i<n;i++){
ans=max(ans,traverse(i,0,n,m,M));
}
return ans;
}
The question says that we can start from any cell in the first column.
As the specific cell is not specified, I have tried finding out the maximum of all possible answers with each cell of the first column as the starting point. But this is giving a TLE, which is expected because the above code would take o(n^2*m) time.
Can anyone help me out with how to get around this problem of optimally figuring out which cell to start from so that the memoization based approach works under the time constraints provided?

Answer 1

Use call by const reference for the STL container. You performed thousands of copies of the full, read-only input.
int dp[55][55];
int traverse(int x,int y,int n,int m, const vector<vector<int>>&M){
. . .
}
int maxGold(int n, int m, const vector<vector<int>>&M)`
{
...
}

Answer 2

here is my solution:
class Solution{
public:
int maxGold(int n, int m, vector<vector<int>> M)
{
vector<vector<int>> f(n + 10, vector<int>(m + 10, 0));
for(int i = 0; i < n; i++){
f[i][0] = M[i][0];
}
for(int j = 1; j < m; j++){
for(int i = 0; i < n; i++){
if(i == 0){
f[i][j] = max(f[i][j-1], f[i+1][j-1]) + M[i][j];
}else if(i == n-1){
f[i][j] = max(f[i][j-1], f[i-1][j-1]) + M[i][j];
}else{
f[i][j] = max({f[i][j-1], f[i-1][j-1], f[i+1][j-1]}) + M[i][j];
}
}
}
int res = 0;
for(int i = 0; i < n; i++){
res = max(res, f[i][m-1]);
}
return res;
}
};