What is the time complexity of 132 Pattern?

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

What data structures are used to solve 132 Pattern?

This problem uses Array, Binary Search, Stack, Monotonic Stack, Ordered Set. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving 132 Pattern?

Insight 1: Scanning the array from right to left allows us to maintain a potential '2' value using a stack. This helps in efficiently finding '1' such that '1 < 3 < 2'. Insight 2: Using a stack (or similar data structure for C) to keep track of potential '2' values allows us to optimize the search for the '3' value. The stack helps us efficiently find the largest value less than the current number (potential '1').

Which companies ask LeetCode 456. 132 Pattern in interviews?

This problem is frequently asked at IBM, Intuit.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

132 Pattern - LeetCode 456 Solution

Q: How to solve 132 Pattern?

The solution utilizes a stack and a variable `s2` (representing the '3' in the 132 pattern) to efficiently find the pattern. It iterates through the array from right to left. The stack stores potential '2' values (the middle number in the 132 pattern). The `s2` variable stores the largest value that has been popped from the stack so far, acting as the potential '3' value. For each number encountered (potential '1'), the solution checks if it's less than `s2`. If it is, a 132 pattern has been found. Otherwise, the stack is updated by popping elements smaller than the current number (because the current number is a better potential '2'), and the current number is then pushed onto the stack.

Q: Which companies ask LeetCode 456. 132 Pattern in interviews?

This problem is frequently asked at IBM, Intuit.

132 Pattern - Complete Solution Guide

132 Pattern is LeetCode problem 456, 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 array of n integers nums , a 132 pattern is a subsequence of three integers nums[i] , nums[j] and nums[k] such that i < j < k and nums[i] < nums[k] < nums[j] . Return true if there is a 132 pattern in nums , otherwise, return false . Example 1: Input: nums = [1,2,3,4] Output: false Explanation: There is no 132 pattern in the sequence. Example 2: Input: nums = [3,1,4,2] Output: true Explanation: There is a 132 pattern in the sequence: [1, 4, 2]. Example 3: Input: nums = [-1,3,2,0] Output

Detailed Explanation

The problem asks us to find a "132 pattern" in a given array of integers. A 132 pattern is a subsequence of three numbers, `nums[i]`, `nums[j]`, and `nums[k]`, where `i < j < k` and `nums[i] < nums[k] < nums[j]`. In simpler terms, we need to find three numbers in the array such that the first number is less than the third number, and the third number is less than the second number, and they appear in that order in the array. The input is an array of integers, and the output is a boolean value: `true` if a 132 pattern exists, and `false` otherwise. The constraints are the size of the array and the range of values within the array.

Solution Approach

The solution utilizes a stack and a variable `s2` (representing the '3' in the 132 pattern) to efficiently find the pattern. It iterates through the array from right to left. The stack stores potential '2' values (the middle number in the 132 pattern). The `s2` variable stores the largest value that has been popped from the stack so far, acting as the potential '3' value. For each number encountered (potential '1'), the solution checks if it's less than `s2`. If it is, a 132 pattern has been found. Otherwise, the stack is updated by popping elements smaller than the current number (because the current number is a better potential '2'), and the current number is then pushed onto the stack.

Step-by-Step Algorithm

Step 1: Initialize an empty stack to store potential '2' values and `s2` to negative infinity (or Integer.MIN_VALUE).
Step 2: Iterate through the input array `nums` from right to left (from index `n-1` to `0`).
Step 3: In each iteration, let `num` be the current element `nums[i]`. Check if `num` is less than `s2`. If so, return `true` (a 132 pattern is found).
Step 4: While the stack is not empty and `num` is greater than the top element of the stack, pop the top element from the stack and update `s2` with the popped value. This ensures `s2` always holds the largest value popped so far, acting as the '3' in the 132 pattern.
Step 5: Push `num` onto the stack. This adds `num` as a potential '2' value for future iterations.
Step 6: After iterating through the entire array without finding a 132 pattern, return `false`.

Key Insights

Insight 1: Scanning the array from right to left allows us to maintain a potential '2' value using a stack. This helps in efficiently finding '1' such that '1 < 3 < 2'.
Insight 2: Using a stack (or similar data structure for C) to keep track of potential '2' values allows us to optimize the search for the '3' value. The stack helps us efficiently find the largest value less than the current number (potential '1').
Insight 3: The s2 variable keeps track of the largest value that has been popped from the stack, which represents a potential '3' value in the 132 pattern. If we find a '1' (nums[i]) that is less than '3' (s2), then we have found a 132 pattern.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search, Stack, Monotonic Stack, Ordered Set.

Companies

Asked at: IBM, Intuit.

132 Pattern | LeetCode 456

Medium

Array Binary Search Stack Monotonic Stack Ordered Set

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

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public boolean find132pattern(int[] nums) {
        int n = nums.length;
        if (n < 3) {
            return false;
        }

        Stack<Integer> stack = new Stack<>();
        int s2 = Integer.MIN_VALUE;

        for (int i = n - 1; i >= 0; i--) {
            int num = nums[i];

            if (num < s2) {
                return true;
            }
            
            while (!stack.isEmpty() && num > stack.peek()) {
                s2 = stack.pop();
            }
            
            stack.push(num);
            
        }
        return false;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 456: 132 Pattern - Detailed explanation and solution approaches.

Topics

Array Binary Search Stack Monotonic Stack Ordered Set

Companies

IBMIntuit

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:456

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

6 steps

1
Step 1: Initialize an empty stack to store potential '2' values and `s2` to negative infinity (or Integer.MIN_VALUE).
2
Step 2: Iterate through the input array `nums` from right to left (from index `n-1` to `0`).
3
Step 3: In each iteration, let `num` be the current element `nums[i]`. Check if `num` is less than `s2`. If so, return `true` (a 132 pattern is found).
4
Step 4: While the stack is not empty and `num` is greater than the top element of the stack, pop the top element from the stack and update `s2` with the popped value. This ensures `s2` always holds the largest value popped so far, acting as the '3' in the 132 pattern.
5
Step 5: Push `num` onto the stack. This adds `num` as a potential '2' value for future iterations.
6
Step 6: After iterating through the entire array without finding a 132 pattern, return `false`.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is derived from the single pass through the array. The stack operations, although nested, contribute linearly because each element is added and removed at most once. Space complexity is driven by the potential size of the stack, which can grow to hold all the input elements in certain scenarios.

📖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

Brute Force

Iterate through all possible combinations of i, j, and k and check if `nums[i] < nums[k] < nums[j]` and `i < j < k`.

Time: O(n^3)Space: O(1)

Trade-offs: This approach is very simple to understand but extremely inefficient for larger arrays. It's suitable only for very small input sizes. The main solution is significantly better in terms of performance.

Using prefix minimums and binary search

Calculate the prefix minimums, then iterate through the array, and for each element `nums[j]`, use binary search on the suffix array after index `j` to find a value `nums[k]` such that `prefixMin[j-1] < nums[k] < nums[j]`.

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

Trade-offs: This is better than brute force but still not as good as the stack-based approach.

✨

Key Insights

1Insight 1: Scanning the array from right to left allows us to maintain a potential '2' value using a stack. This helps in efficiently finding '1' such that '1 < 3 < 2'.
2Insight 2: Using a stack (or similar data structure for C) to keep track of potential '2' values allows us to optimize the search for the '3' value. The stack helps us efficiently find the largest value less than the current number (potential '1').
3Insight 3: The s2 variable keeps track of the largest value that has been popped from the stack, which represents a potential '3' value in the 132 pattern. If we find a '1' (nums[i]) that is less than '3' (s2), then we have found a 132 pattern.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Edge Cases

Edge case 1: Empty or very small array (n < 3). The code explicitly checks for this and returns `false` because a 132 pattern requires at least three elements.
Edge case 2: Sorted arrays (ascending or descending). The algorithm correctly identifies that there are no 132 patterns in such arrays. In an ascending sorted array, the stack will remain empty or contain only the last element, never satisfying the `num < s2` condition. In a descending array, the `while` loop will continuously pop elements from the stack, ensuring `s2` is always the largest value seen so far, and again, `num < s2` will never be true.
Edge case 3: Array with repeating elements. The algorithm handles this correctly because it always looks for a '1' value that is less than the '3' (s2) regardless of how many of the same value are in the array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not iterating from right to left. Iterating from left to right makes it much harder to maintain the required conditions efficiently.

✗Common mistake 2: Incorrectly updating `s2`. It's important to update `s2` with the value popped from the stack *before* pushing the current number onto the stack to ensure `s2` represents the largest number to the right of the current number.

✗Common mistake 3: Misunderstanding the role of the stack. The stack is used to find a potential '2' that is greater than '3' (s2). The order in the stack must be monotonically decreasing (from top to bottom).

✅

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 patterns in stock prices that could indicate potential trading opportunities. For example, a 132 pattern could suggest a potential short-selling opportunity.

🏦Real-world scenario 2: Signal processing. Detecting specific signal patterns in a stream of data. The '132 pattern' could be adapted to identify a specific sequence of events or features in the signal.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: September 5, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

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 avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

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 "132 Pattern"?

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 binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?