LeetCode#334.

Increasing Triplet Subsequence

Problem Statement

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?

Golang Solution Increasing Triplet Subsequence

func increasingTriplet(nums []int) bool {
	n := len(nums)
	if (n < 3) || (n > (5 * 10 * 10 * 10 * 10 * 10)) {
		return false
	}

	first := math.MaxInt64
	second := math.MaxInt64

	for i := 0; i < n; i++ {
		if nums[i] <= first {
			first = nums[i]
		} else if nums[i] <= second {
			second = nums[i]
		} else {
			return true
		}
	}
	return false
}

Explanation

• first stores the smallest value seen so far.
• second stores the second smallest value.
• As we traverse the array, we update first and second if possible. If we find any number greater than both, it confirms the existence of an increasing triplet subsequence.

Time Complexity

• O(n) since we are iterating through the array once.

Space Complexity

• O(1) since we only use a few variables to track the smallest values.

This approach is optimal and works efficiently for large input arrays.

If you’d like to see another solution, like dynamic programming or a different approach, let me know!

Visit https: https://codeandalgo.com for more such contents