# c++ O(1) space O(n) time single pass solution 23ms

• All you need to do is to maintain two interval's edge.
One is [current min, max from right of current min]. The other is [min from left of current max, current max]. Andy number falls into one of these interval, we return true.
It is not very tricky to maintain. You could understand it easily from the code below.

bool find132pattern(vector<int>& nums) {
if (nums.size() < 3)
return false;
int curmin, curminMax, curmax, curmaxMin;
curmin = curminMax = curmax = curmaxMin = nums[0];
for (int i = 1; i < nums.size(); ++i) {
if (nums[i] < curmin) {
curmin = curminMax = nums[i];
} else if (nums[i] > curmax) {
curmax = curminMax = nums[i];
curmaxMin = curmin;
} else {
if (nums[i] > curmin && curminMax > nums[i])
return true;
if (nums[i] > curmaxMin && curmax > nums[i])
return true;
if (nums[i] > curminMax)
curminMax = nums[i];
}
}
return false;
}

• Try [90, 100, 50, 70, 20, 40, 10, 51], it should return true but your code returned false.

It seems that there isn't an algorithm which can solve the problem in O(n) time.

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