What is the time complexity of Increasing Triplet Subsequence?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Increasing Triplet Subsequence?

This problem uses Array, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Increasing Triplet Subsequence?

Insight 1: We don't need to find the actual indices; we only need to determine if such a triplet exists. Insight 2: Maintaining the smallest and second smallest values seen so far allows us to efficiently check if a larger value exists that completes the triplet.

Which companies ask LeetCode 334. Increasing Triplet Subsequence in interviews?

This problem is frequently asked at Coupang, FactSet, IBM, MakeMyTrip, Nutanix.

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Increasing Triplet Subsequence - LeetCode 334 Solution

Q: How to solve Increasing Triplet Subsequence?

The provided solution uses a greedy approach to efficiently determine the existence of an increasing triplet subsequence. It iterates through the array, maintaining two variables, `first` and `second`, representing the smallest and second smallest numbers encountered so far, respectively. For each number in the array, the algorithm compares it against `first` and `second`. If a number is smaller than or equal to `first`, we update `first`. If a number is greater than `first` but smaller than or equal to `second`, we update `second`. If a number is greater than both `first` and `second`, it means we've found an increasing triplet, and we return `true`. If the loop completes without finding such a triplet, we return `false`.

Increasing Triplet Subsequence - Complete Solution Guide

Increasing Triplet Subsequence is LeetCode problem 334, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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: One of the valid triplet is (3, 4, 5), because nums[3] == 0 < nums[

Detailed Explanation

The problem asks us to determine whether an integer array `nums` contains an increasing subsequence of length 3. More formally, we need to find indices `i`, `j`, and `k` such that `i < j < k` and `nums[i] < nums[j] < nums[k]`. If such a subsequence exists, we return `true`; otherwise, we return `false`. The constraints indicate that the array's length can be up to 5 * 10^5, and the integer values are within the standard 32-bit range. The follow-up challenges us to solve this in O(n) time and O(1) space.

Solution Approach

The provided solution uses a greedy approach to efficiently determine the existence of an increasing triplet subsequence. It iterates through the array, maintaining two variables, `first` and `second`, representing the smallest and second smallest numbers encountered so far, respectively. For each number in the array, the algorithm compares it against `first` and `second`. If a number is smaller than or equal to `first`, we update `first`. If a number is greater than `first` but smaller than or equal to `second`, we update `second`. If a number is greater than both `first` and `second`, it means we've found an increasing triplet, and we return `true`. If the loop completes without finding such a triplet, we return `false`.

Step-by-Step Algorithm

Step 1: Initialize `first` and `second` to positive infinity (or the maximum possible integer value in the given language).
Step 2: Iterate through the `nums` array.
Step 3: For each number `num` in `nums`:
Step 4: If `num` is less than or equal to `first`, update `first` to `num`. This ensures `first` always holds the smallest value seen so far.
Step 5: Else if `num` is less than or equal to `second`, update `second` to `num`. This means `num` is greater than `first` but not greater than `second`.
Step 6: Else (if `num` is greater than both `first` and `second`), return `true`. This is because we have found a third number that is greater than both `first` and `second`, completing the increasing triplet.
Step 7: If the loop completes without returning `true`, return `false`. This means no increasing triplet was found in the array.

Key Insights

Insight 1: We don't need to find the actual indices; we only need to determine if such a triplet exists.
Insight 2: Maintaining the smallest and second smallest values seen so far allows us to efficiently check if a larger value exists that completes the triplet.
Insight 3: The key to achieving O(1) space complexity is to avoid storing the entire subsequence; instead, keep track of the smallest two values seen so far.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy.

Companies

Asked at: Coupang, FactSet, IBM, MakeMyTrip, Nutanix.

Increasing Triplet Subsequence | LeetCode 334

Medium

Array Greedy

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean increasingTriplet(int[] nums) {
        int first = Integer.MAX_VALUE;
        int second = Integer.MAX_VALUE;

        for (int num : nums) {
            if (num <= first) {
                first = num;
            } else if (num <= second) {
                second = num;
            } else {
                return true;
            }
        }

        return false;
    }
}

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Problem Info

Description

Problem 334: Increasing Triplet Subsequence - Detailed explanation and solution approaches.

Topics

Array Greedy

Companies

CoupangFactSetIBMMakeMyTripNutanix

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:334

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize `first` and `second` to positive infinity (or the maximum possible integer value in the given language).
2
Step 2: Iterate through the `nums` array.
3
Step 3: For each number `num` in `nums`:
4
Step 4: If `num` is less than or equal to `first`, update `first` to `num`. This ensures `first` always holds the smallest value seen so far.
5
Step 5: Else if `num` is less than or equal to `second`, update `second` to `num`. This means `num` is greater than `first` but not greater than `second`.
6
Step 6: Else (if `num` is greater than both `first` and `second`), return `true`. This is because we have found a third number that is greater than both `first` and `second`, completing the increasing triplet.
7
Step 7: If the loop completes without returning `true`, return `false`. This means no increasing triplet was found in the array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through the array only once (O(n)). It uses only two integer variables (`first` and `second`) to store the smallest two numbers encountered so far, regardless of the input size. Therefore, the space complexity is constant (O(1)).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

Use DP to build a table `dp[i][j]` representing whether there exists a subsequence of length `j+1` ending at index `i`. However, the space complexity will not be O(1).

Time: O(n^2)Space: O(n^2)

Trade-offs: This approach is less efficient in terms of both time and space, making it unsuitable when O(1) space complexity is required.

Binary Search (similar to Longest Increasing Subsequence)

Similar to the longest increasing subsequence problem, we can maintain a list of the smallest tail elements of all increasing subsequences of length 1, 2, and 3. If a new element is greater than all existing tail elements, we've found an increasing triplet.

Time: O(n log n)Space: O(1)

Trade-offs: While this provides O(n log n) time, the greedy approach is more optimal since it provides O(n) time with constant space.

✨

Key Insights

1Insight 1: We don't need to find the actual indices; we only need to determine if such a triplet exists.
2Insight 2: Maintaining the smallest and second smallest values seen so far allows us to efficiently check if a larger value exists that completes the triplet.
3Insight 3: The key to achieving O(1) space complexity is to avoid storing the entire subsequence; instead, keep track of the smallest two values seen so far.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: Empty array or array with fewer than three elements. The algorithm correctly returns `false` in these cases, as no triplet can be formed.
Edge case 2: Array with all elements equal. The algorithm correctly returns `false` because no increasing subsequence exists.
Edge case 3: Array with decreasing sequence. The algorithm correctly returns `false`.
Edge case 4: Array with a valid triplet at the beginning or end. The algorithm correctly identifies the triplet.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly assuming that the first two smallest numbers encountered will always be part of the increasing triplet. The `first` and `second` variables are merely placeholders; a smaller number encountered later can potentially lead to a better increasing subsequence.

✗Common mistake 2: Attempting to store the entire increasing subsequence, leading to O(n) space complexity.

✗Common mistake 3: Using a brute-force approach with three nested loops, resulting in O(n^3) time complexity, which is inefficient for large input arrays.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Stock market analysis. Identifying trends in stock prices where a price increases over three consecutive days (or any similar pattern).

🏦Real-world scenario 2: Data analysis. Detecting patterns in sensor data where a measurement increases over a specific duration.

🎮Real-world scenario 3: Signal processing. Identifying increasing patterns in audio or video signals.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 2, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Increasing Triplet Subsequence"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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