What is the brute force approach and complexity for 132 Pattern?

Three nested loops over i < j < k with time complexity O(N³) and space complexity O(1).

What is the optimal approach and complexity for 132 Pattern?

Scan right to left; values popped by a bigger value become the "2"; keep the largest with optimal time complexity O(N) and space complexity O(N).

How do you identify the pattern for 132 Pattern?

Look for key signals in the prompt: "i < j < k with nums[i] < nums[k] < nums[j]" indicates Three positions with an order and a value condition. Fix the hardest part (the "3, 2" pair) with a stack, the easy part (the "1") with a single comparison.; "nums[k] < nums[j], j before k" indicates The "2" needs a bigger value on its left — a previous-greater relation, which a monotonic stack finds when it pops.; "return true / false" indicates We only need one pattern, so we keep only the most useful candidate (the largest "2")..

132 Pattern

Q: How do you identify the pattern for 132 Pattern?

Look for key signals in the prompt: "i < j < k with nums[i] < nums[k] < nums[j]" indicates Three positions with an order and a value condition. Fix the hardest part (the "3, 2" pair) with a stack, the easy part (the "1") with a single comparison.; "nums[k] < nums[j], j before k" indicates The "2" needs a bigger value on its left — a previous-greater relation, which a monotonic stack finds when it pops.; "return true / false" indicates We only need one pattern, so we keep only the most useful candidate (the largest "2")..

Medium·LeetCode #456 Pro OnlyTricky / Edge Case

LeetCode

Problem Statement

You are given an integer array nums.

A 132 pattern is three positions i < j < k where

nums[i] < nums[k] < nums[j].

The name comes from the shape: the first value is the smallest ("1"), the middle one is the largest ("3"), and the last one is in between ("2"). The three positions don't have to be next to each other.

Return true if the array contains a 132 pattern, otherwise false.

Input

An integer array nums.

Output

true or false.

Example 1

Input: nums = [1, 2, 3, 4]

Output: false

The values only go up, so no later value is ever smaller than an earlier one.

Example 2

Input: nums = [3, 1, 4, 2]

Output: true

1, 4, 2 is a 132 pattern: 1 < 2 < 4.

Example 3

Input: nums = [-1, 3, 2, 0]

Output: true

There are three: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].

Constraints

• 1 ≤ n ≤ 2·10^5
• −10^9 ≤ nums[i] ≤ 10^9

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“i < j < k with nums[i] < nums[k] < nums[j]”

→ Three positions with an order and a value condition. Fix the hardest part (the "3, 2" pair) with a stack, the easy part (the "1") with a single comparison.

“nums[k] < nums[j], j before k”

→ The "2" needs a bigger value on its left — a previous-greater relation, which a monotonic stack finds when it pops.

“return true / false”

→ We only need one pattern, so we keep only the most useful candidate (the largest "2").

Approach Complexity Ladder

Brute Force — try every tripleO(N³)O(1)

Three nested loops over i < j < k

Better — prefix minimum as the "1"O(N²)O(1)

For each j, use the smallest value before it as the "1", then scan k > j

Optimal — right-to-left stack with the best "2"O(N)O(N)

Scan right to left; values popped by a bigger value become the "2"; keep the largest

Brute Force Intuition

Try every i < j < k and check nums[i] < nums[k] < nums[j]. Return true on the first match.

Recognition rule for this sub-pattern:

"How many consecutive previous items are ≤ me?" Popped items fold their span into the new one.

Remember this · Monotonic Stack

Trigger

Next/previous greater or smaller element, spans, histogram areas, "for each element, how far does it dominate?"

Think

When a new element arrives, which older elements can never be the answer again?

Key question

Is my stack increasing or decreasing, and what do I record when I pop?

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