334. Increasing Triplet Subsequence – Leetcode Solutions

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.

class Solution:
    def increasingTriplet(self, nums: List[int]) -> bool:
        # Initialize two variables to track the smallest and second smallest numbers
        first = float('inf')
        second = float('inf')
        
        # Traverse through the array
        for num in nums:
            if num <= first:
                # If current number is smaller than or equal to 'first', update 'first'
                first = num
            elif num <= second:
                # If current number is greater than 'first' but smaller than or equal to 'second', update 'second'
                second = num
            else:
                # If we find a number larger than both 'first' and 'second', return True
                return True
        
        # If no increasing triplet is found, return False
        return False

class Solution {
public:
    bool increasingTriplet(std::vector<int>& nums) {
        // Edge case: If the array has less than 3 elements, we cannot have a triplet
        if (nums.size() < 3) {
            return false;
        }
        
        // Initialize two variables with large values
        int first = INT_MAX, second = INT_MAX;
        
        // Traverse through the array
        for (int num : nums) {
            // If the current number is smaller than or equal to the first, update the first
            if (num <= first) {
                first = num;
            }
            // If the current number is greater than first but smaller than or equal to the second, update the second
            else if (num <= second) {
                second = num;
            }
            // If the current number is greater than both first and second, we found a triplet
            else {
                return true;
            }
        }
        
        // No increasing triplet was found
        return false;
    }
};

var increasingTriplet = function(nums) {
    // Initialize two variables, first and second, to hold the smallest and second smallest values we've encountered.
    let first = Infinity;
    let second = Infinity;
    
    // Loop through the array of numbers
    for (let num of nums) {
        // Check if the current number is smaller than or equal to the 'first' variable
        // If so, update 'first' to the current number. This means we're still looking for the smallest number.
        if (num <= first) {
            first = num;
        } 
        // Otherwise, check if the current number is smaller than or equal to 'second'
        // and greater than 'first'. If so, update 'second' to the current number. This means
        // we're looking for the second number in the increasing triplet.
        else if (num <= second) {
            second = num;
        } 
        // If the current number is greater than both 'first' and 'second', then we have found the third number
        // in the increasing triplet and can return true.
        else {
            return true;
        }
    }

    // If we complete the loop without finding a valid increasing triplet, return false.
    return false;
};

class Solution {
    public boolean increasingTriplet(int[] nums) {
        // Step 1: Check for array length. If it's less than 3, it's impossible to find a triplet
        if (nums.length < 3) {
            return false;
        }

        // Step 2: Initialize two variables to track the first and second smallest elements
        int first = Integer.MAX_VALUE;  // This will track the smallest number
        int second = Integer.MAX_VALUE; // This will track the second smallest number

        // Step 3: Traverse through the array and attempt to find an increasing triplet
        for (int num : nums) {
            // If the current number is smaller than or equal to 'first', update 'first'
            if (num <= first) {
                first = num;
            } 
            // If the current number is smaller than or equal to 'second' but larger than 'first', update 'second'
            else if (num <= second) {
                second = num;
            } 
            // If we find a number greater than both 'first' and 'second', we have found a valid triplet
            else {
                return true;
            }
        }

        // Step 4: If we complete the loop without finding such a triplet, return false
        return false;
    }
}