334. Increasing Triplet Subsequence

334. Increasing Triplet Subsequence | LeetCode Solution

Given an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.

Example 1:

Input: nums = [1,2,3,4,5]
Output: true
Explanation: Any triplet where i < j < k is valid.

Example 2:

Input: nums = [5,4,3,2,1]
Output: false
Explanation: No triplet exists.

Example 3:

Input: nums = [2,1,5,0,4,6]
Output: true
Explanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6.

Constraints:

1 <= nums.length <= 5 * 105
-231 <= nums[i] <= 231 - 1

Follow up: Could you implement a solution that runs in O(n) time complexity and O(1) space complexity?

class Solution
{
public:
  bool increasingTriplet(vector<int> &nums)
  {
    vector<int> next(nums.size(), -1);
    vector<int> prev(nums.size(), -1);
    stack<int> st;
    for (int i = 0; i < nums.size(); i++)
    {
      while (!st.empty() && nums[st.top()] < nums[i])
      {
        next[st.top()] = i;
        st.pop();
      }
      st.push(i);
    }
    while (!st.empty())
      st.pop();

    for (int i = nums.size() - 1; i >= 0; i--)
    {
      while (!st.empty() && nums[st.top()] > nums[i])
      {
        prev[st.top()] = i;
        st.pop();
      }
      st.push(i);
    }

    for (int i = 0; i < nums.size(); i++)
      if (prev[i] != -1 && next[i] != -1)
        return true;
    return false;
  }
};

class Solution
{
public:
  bool increasingTriplet(vector<int> &nums)
  {
    int firstMin = INT_MAX;
    int secondMin = INT_MAX;

    for (int i : nums)
    {
      if (i <= firstMin)
      {
        firstMin = i;
      }
      else if (i <= secondMin)
      {
        secondMin = i;
      }
      else
        return true;
    }
    return false;
  }
};