The problem can be found at the following link: Question Link
We need to check if there exists three indices i,j and k such that i < j < k and nums[i] < nums[k] < nums[j]. To do this we'll have to first find if there exists a pair k, j for which k > j and at the same time nums[j] > nums[k]. Once this pair is found we need to find if we have a number to the left of j whose value is less than nums[k]. We can use monotonic stack for this.
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Start with making our s3 = INT_MIN.
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Initialise an empty stack st. 
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Now starting from our last index till the first index, i.e from i = n- 1 to i = 0, if our current number is lesser than s3 i.enums[i] < s3, we've found our s1 we return true because we have already found a number s2 such thats3 < s2so our sequences1(nums[i]) < s3 < s2is complete.
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Else if nums[i] > s3it means nums[i] could be our potential s2, so , while our nums[i] > the top of our stack, we pop and update the value of s3. The popping and updating is done because we need our s3 to be smaller than nums[i] sinces3 < s2.
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Push nums[i]into the stack.
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At last return false, it would mean that our true isn't returned so no such sequence exists. 
- Time Complexity: O(n)
- Auxiliary Space Complexity: O(n)
class Solution {
public:
    bool find132pattern(vector<int>& nums) {
        int n = nums.size();
        int s3 = INT_MIN;
        stack<int> st;
        for(int i=n-1; i>=0; i--)
        {
            if(nums[i] < s3) return true;
            else
            {
                while(!st.empty() && nums[i] > st.top())
                {
                    s3 = st.top();
                    st.pop();
                }
                st.push(nums[i]);
            }
        }
        return false;
    }
};
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