334 Increasing Triplet Subsequence 1

Joel Castillo Espinosa 2

DESCRIPTION

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.

Examples

  • 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 ≤ $5*10^5$
  • $-2^{31}$ ≤ nums[i] ≤ $2^{31}$ - 1

SOLUTION 3

triple<- function(nums) {
  first <- Inf
  second <- Inf
  
  for (i in 1:length(nums)){
    # is nums[i] less than first element in subsequence?
    if(nums[i] <= first){
      first <- nums[i]
    #  is nums[i] less than first element in subsequence?
    } else if(nums[i] <= second){
      second <- nums[i]
    # then nums[i] is greater than first and second element in subsequence
    } else {return(TRUE)}
  }
  return(FALSE)
}

Examples using the function

We can use the examples presented before.

triple(c(1,2,3,4,5))

## [1] TRUE

triple(nums = c(5,4,3,2,1))

## [1] FALSE

triple(c(2,1,5,0,4,6))

## [1] TRUE

  1. This problem is originally from LeetCode, you can find it in Leetcode. ↩

  2. Email: jocastillo@colmex.mx. For more content visit my website: https://joelcastillo.netlify.app
    If you have any questions or suggestions, I’d be grateful to hear from you. ↩

  3. This solution is entirely my own work. It was developed using R version 4.4.1 (2024-06-14 ucrt). ↩