# Why my two versions of code so slow One TLE and One inefficient.

• Hi all, I use DFS for this problem, but the time efficiency is too low. In the version one I used unordered_set, which is equivalent to HashSet in Java, to check if a cell has been used for matching. The code can pass all tests, but only fall in the bottom 10% in time efficiency. The user defined stack rather than recursion stack is used in my code.

``````#include <unordered_set>
class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
if (word=="")
return true;
int M = board.size(), N = board[0].size(), L = word.length();
int temp_i, temp_j, dir, l;
const int up=0, left=1, down=2, right=3, gone=4;
if ((M*N==0)||(M*N<L))
return false;
stack<pair<int,int> > st;
unordered_set<int> forbidden;
// first = i*N+j, second = direction;
// 0: up, 1: left, 2: down, 3: right
for (int i=0; i<M; i++)
for (int j=0; j<N; j++)
{
if (board[i][j]==word[0])
{
if (L==1)
return true;
st.push(make_pair(i*N+j,up));
forbidden.insert(i*N+j);
while (!st.empty())
{
if (forbidden.size()==L)
return true;
temp_i = st.top().first; dir = st.top().second;
temp_j = temp_i%N; temp_i /= N;
if (dir==up)  // to check up
{
st.pop();
st.push(make_pair(temp_i*N+temp_j,right));
if ((temp_i==0)||(forbidden.count((temp_i-1)*N+temp_j)==1)||(board[temp_i-1][temp_j]!=word[forbidden.size()]))
continue;
forbidden.insert((temp_i-1)*N+temp_j);
st.push(make_pair((temp_i-1)*N+temp_j,up));
continue;
}
if (dir==right)
{
st.pop();
st.push(make_pair(temp_i*N+temp_j,down));
if ((temp_j==N-1)||(forbidden.count(temp_i*N+temp_j+1)==1)||(board[temp_i][temp_j+1]!=word[forbidden.size()]))
continue;
forbidden.insert(temp_i*N+temp_j+1);
st.push(make_pair(temp_i*N+temp_j+1,up));
}
if (dir==down)  // to check up
{
st.pop();
st.push(make_pair(temp_i*N+temp_j,left));
if ((temp_i==M-1)||(forbidden.count((temp_i+1)*N+temp_j)==1)||(board[temp_i+1][temp_j]!=word[forbidden.size()]))
continue;
forbidden.insert((temp_i+1)*N+temp_j);
st.push(make_pair((temp_i+1)*N+temp_j,right));
continue;
}
if (dir==left)
{
st.pop();
st.push(make_pair(temp_i*N+temp_j,gone));
if ((temp_j==0)||(forbidden.count(temp_i*N+temp_j-1)==1)||(board[temp_i][temp_j-1]!=word[forbidden.size()]))
continue;
forbidden.insert(temp_i*N+temp_j-1);
st.push(make_pair(temp_i*N+temp_j-1,up));
}
if (dir==gone)
{
st.pop();
forbidden.erase(temp_i*N+temp_j);
}
}
}
}
return false;
}
};
``````

Then I switch from the unordered_set to a 2-D bool array for availability checking, hope that extra space can buy me some time. However, it got TLE this time on the case ["aaa...aa",...,"aaa...ab"], "aa...ab".
Can some one tell me what makes my two codes inefficient? thanks!
I don't think there is problem in correctness, since it is modified from my AC code.

``````class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
// This part is the same as version one
stack<pair<int,int> > st;
bool **Used = new bool*[M];
for (int i=0; i<M; i++)
Used[i] = new bool [N];
for (int i=0; i<M; i++)
for (int j=0; j<N; j++)
{
if (board[i][j]==word[0])
{
if (L==1)
return true;
st.push(make_pair(i*N+j,up));
Used[i][j] = true;
l = 1;
while (!st.empty())
{
if (l==L)
return true;
temp_i = st.top().first; dir = st.top().second;
temp_j = temp_i%N; temp_i /= N;
if (dir==up)  // to check up
{
st.pop();
st.push(make_pair(temp_i*N+temp_j,right));
if ((temp_i==0)||(Used[temp_i-1][temp_j])||(board[temp_i-1][temp_j]!=word[l]))
continue;
Used[temp_i-1][temp_j] = true;
l++;
st.push(make_pair((temp_i-1)*N+temp_j,up));
continue;
}
// EACH POST HAS 8000 CHAR LIMIT, SO THE RIGHT TO LEFT PART IS OMITTED HERE
if (dir==gone)
{
st.pop();
Used[temp_i][temp_j] = false;
l--;
}
}
}
}
for (int i=0; i<M; i++)
delete [] Used[i];
delete [] Used;
return false;
}
};``````

• You can use `board` to record which cell is traversed, restore it's value if need backtrace.
This will save program's running time.

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