text: // Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.
  
  // Formally the function should:
  // Return true if there exists i, j, k
  // such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.
  // Your algorithm should run in O(n) time complexity and O(1) space complexity.
  
  // Examples:
  // Given [1, 2, 3, 4, 5],
  // return true.
  
  // Given [5, 4, 3, 2, 1],
  // return false.
  
  // Credits:
  // Special thanks to @DjangoUnchained for adding this problem and creating all test cases.
  
  // Hide Company Tags Facebook
  // Hide Similar Problems (M) Longest Increasing Subsequence
  
  /**
   * @param {number[]} nums
   * @return {boolean}
   */
  var increasingTriplet = function (nums) {
    var min1 = Infinity;
    var min2 = Infinity;
  
    for (var i = 0; i < nums.length; i++) {
      if (nums[i] <= min1) {
        min1 = nums[i];
      } else if (nums[i] <= min2) {
        min2 = nums[i];
      } else {
        return true;
      }
    }
  
    return false;
  };
  