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 exist, 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 * 10^5
• -2^31 <= nums[i] <= 2^31 - 1

Follow-up:

Can you implement a solution that runs in O(n) time complexity and O(1) space complexity?

Initial Thought Process:

To solve this problem efficiently, we need to keep track of the smallest and second smallest values encountered so far. If we find a third value that is greater than the second smallest, then we have found an increasing triplet.

Basic Solution (Brute Force):

The brute force solution involves checking all possible triplets to see if there exists one that satisfies the condition i < j < k and nums[i] < nums[j] < nums[k]. This approach has a time complexity of O(n^3), which is not efficient for large input sizes.

Code:

function increasingTripletBruteForce(nums: number[]): boolean {
    const n = nums.length;
    for (let i = 0; i < n - 2; i++) {
        for (let j = i + 1; j < n - 1; j++) {
            for (let k = j + 1; k < n; k++) {
                if (nums[i] < nums[j] && nums[j] < nums[k]) {
                    return true;
                }
            }
        }
    }
    return false;
}

Time Complexity Analysis:

• Time Complexity: O(n^3), where n is the length of the array. This is because we are checking all possible triplets.
• Space Complexity: O(1), as we are not using any extra space.

Limitations:

The brute force solution is not efficient and is not suitable for large input sizes.

Optimized Solution:

The optimized solution involves iterating through the array while maintaining two variables, first and second, which represent the smallest and second smallest values encountered so far. If we find a value greater than second, then we return true.

Code:

function increasingTriplet(nums: number[]): boolean {
    let first = Infinity;
    let second = Infinity;

    for (let num of nums) {
        if (num <= first) {
            first = num; // smallest value
        } else if (num <= second) {
            second = num; // second smallest value
        } else {
            return true; // found a value greater than second smallest, thus an increasing triplet exists
        }
    }

    return false;
}

Time Complexity Analysis:

• Time Complexity: O(n), where n is the length of the array. We iterate through the array once.
• Space Complexity: O(1), as we are using only a constant amount of extra space.

Improvements Over Basic Solution:

• This solution runs in linear time and uses constant space, making it optimal for the given constraints.

Edge Cases and Testing:

Edge Cases:

1. The array is in decreasing order.
2. The array contains exactly three elements in increasing order.
3. The array has a large number of elements with no increasing triplet.
4. The array contains duplicates.

Test Cases:

console.log(increasingTripletBruteForce([1,2,3,4,5])); // true
console.log(increasingTripletBruteForce([5,4,3,2,1])); // false
console.log(increasingTripletBruteForce([2,1,5,0,4,6])); // true
console.log(increasingTripletBruteForce([1,1,1,1,1])); // false
console.log(increasingTripletBruteForce([1,2])); // false
console.log(increasingTripletBruteForce([1,2,3])); // true
console.log(increasingTripletBruteForce([1,5,0,4,1,3])); // true

console.log(increasingTriplet([1,2,3,4,5])); // true
console.log(increasingTriplet([5,4,3,2,1])); // false
console.log(increasingTriplet([2,1,5,0,4,6])); // true
console.log(increasingTriplet([1,1,1,1,1])); // false
console.log(increasingTriplet([1,2])); // false
console.log(increasingTriplet([1,2,3])); // true
console.log(increasingTriplet([1,5,0,4,1,3])); // true

General Problem-Solving Strategies:

1. Understand the Problem: Carefully read the problem statement to understand the requirements and constraints.
2. Identify Key Operations: Determine the key operations needed, such as tracking the smallest and second smallest values.
3. Optimize for Efficiency: Use efficient algorithms and data structures to minimize time and space complexity.
4. Test Thoroughly: Test the solution with various cases, including edge cases, to ensure correctness.

Identifying Similar Problems:

1. Subarray Problems:

   • Problems where you need to find subarrays with specific properties.
   • Example: Finding the maximum sum subarray (Kadane's Algorithm).

2. Two-Pointer Technique:

   • Problems where using two pointers can help optimize the solution.
   • Example: Removing duplicates from a sorted array.

3. In-Place Algorithms:

   • Problems where operations need to be performed in place with limited extra space.
   • Example: Rotating an array to the right by k steps.

Conclusion:

• The problem of finding an increasing triplet subsequence can be efficiently solved using both a brute force approach and an optimized solution with linear time and constant space complexity.
• Understanding the problem and breaking it down into manageable parts is crucial.
• Using efficient algorithms ensures the solution is optimal for large inputs.
• Testing with various edge cases ensures robustness.
• Recognizing patterns in problems can help apply similar solutions to other challenges.

By practicing such problems and strategies, you can improve your problem-solving skills and be better prepared for various coding challenges.