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Program to find a triplet nums[i] < nums[k] < nums[j] from a list nums in C++

Suppose we have a list of numbers called nums, we have to check whether there are triplets (i, j, k) such that i < j < k and nums[i] < nums[k] < nums[j].

So, if the input is like nums = [2, 12, 1, 4, 4], then the output will be True, as [2, 12, 4] matches the criteria because 2 < 4 < 12.

To solve this, we will follow these steps −

  • n := size of nums

  • Define an array left of size n

  • left[0] := nums[0]

  • for initialize i := 1, when i < n, update (increase i by 1), do −

    • left[i] := minimum of nums[i] and left[i - 1]

  • Define one stack st

  • for initialize i := n - 1, when i >= 1, update (decrease i by 1), do −

    • x := left[i - 1]

    • while st is not empty and top of st <= x, do −

      • pop from st

    • if st is not empty and x < nums[i] and nums[i] > top of st, then −

      • return true

    • push nums[i] into st

  • return false

Example  

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   bool solve(vector<int>& nums) {
      int n = nums.size();
      vector<int> left(n);
      left[0] = nums[0];
      for (int i = 1; i < n; i++) {
         left[i] = min(nums[i], left[i - 1]);
      }
      stack<int> st;
      for (int i = n - 1; i >= 1; i--) {
         int x = left[i - 1];
         while (!st.empty() && st.top() <= x)
            st.pop();
         if (!st.empty() && x < nums[i] && nums[i] > st.top())
            return true;
         st.push(nums[i]);
      }
      return false;
   }
};
bool solve(vector<int>& nums) {
   return (new Solution())->solve(nums);
}
int main(){
   vector<int> v = {2, 12, 1, 4, 4};
   cout << solve(v);
}

Input

{2, 12, 1, 4, 4}

Output

1