# Time Limit Exceeded on submission, but passed on Run code

• I wrote a recursive solution which is not of constant space complexity but I hoped it will be fast enough to be passed.

``````class Solution {
bool verifyPreorder(int left, int right, vector<int> const& preorder) {
if(left >= right) { return true; }

int root = preorder[left];
int leftIndx = left + 1, rightIndx = right + 1;
for(int k = left + 1; k <= right; ++k) {
if(preorder[k] > root) {
rightIndx = k;
break;
}
}
for(int k = rightIndx; k <= right; ++k) {
if(preorder[k] < root) {
return false;
}
}

return verifyPreorder(left + 1, rightIndx - 1, preorder) and verifyPreorder(rightIndx, right, preorder);
}

public:
bool verifyPreorder(vector<int>& preorder) {
if(preorder.empty()) return true;
return verifyPreorder(0, (int)preorder.size() - 1, preorder);
}
};
``````

But when I submitted pressing `Submit Solution`, the OJ yielded Time Limit Exceeded for below test case (brief):

``````[8000,7999,7998,7997,7996,7995,7994,7993,7992,7991,7990,7989,7988,7987,7986,7985,7984,7983,7982,7981,7980,7979,7978,............................. ....... .........,13,12,11,10,9,8,7,6,5,4,3,2,1]
``````

Then I tried this as custom testcase and the code ran within time and took 152ms. Is this time too much? Is the tester program different for OJ and testing code?

• It seems the recursion based solution was not fast enough for OJ. So I took some idea from some threads and ended up writing stack based `O(n)` space solution and another constant space solution.

``````class Solution {
bool verifyPreorder(int left, int right, vector<int> const& preorder) {
if(left >= right) { return true; }

int root = preorder[left];
int leftIndx = left + 1, rightIndx = right + 1;
for(int k = left + 1; k <= right; ++k) {
if(preorder[k] > root) {
rightIndx = k;
break;
}
}
for(int k = rightIndx; k <= right; ++k) {
if(preorder[k] < root) {
return false;
}
}

return verifyPreorder(left + 1, rightIndx - 1, preorder) and verifyPreorder(rightIndx, right, preorder);
}

public:
bool verifyPreorder(vector<int>& preorder) {
if(preorder.empty()) return true;

// TLE
// return verifyPreorder(0, (int)preorder.size() - 1, preorder);

// O(n) stack based solution
/*
int Min = INT_MIN;
int n = (int)preorder.size();
stack<int> stk;
stk.push(preorder[0]);

for(int i = 1; i < n; ++i) {
if(preorder[i] < Min) return false;
while(!stk.empty() and preorder[i] > stk.top()) {
Min = stk.top();
stk.pop();
}
stk.push(preorder[i]);
}

return true;
*/

int Min = INT_MIN;
int n = (int)preorder.size();
int i = 0, minIndx = -1;
for(i = 1; i < n and preorder[i] > Min; ++i) {
if(preorder[i] > preorder[i - 1]) { // we entered in right branch
int k = i - 1;
for(int j = i - 2; j > minIndx; --j) {
if(preorder[j] < preorder[i] and preorder[j] > preorder[k]) {
k = j;
}
}
minIndx = k;
Min = preorder[k];
}
}
return (i >= n);
}
};``````

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