LeetCode #456 — MEDIUM

132 Pattern

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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: true
Explanation: There are three 132 patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].

Constraints:

n == nums.length
1 <= n <= 2 * 10⁵
-10⁹ <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Stack · Segment Tree

Example 1

[1,2,3,4]

Example 2

[3,1,4,2]

Example 3

[-1,3,2,0]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #456: 132 Pattern
class Solution {
    public boolean find132pattern(int[] nums) {
        int vk = -(1 << 30);
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = nums.length - 1; i >= 0; --i) {
            if (nums[i] < vk) {
                return true;
            }
            while (!stk.isEmpty() && stk.peek() < nums[i]) {
                vk = stk.pop();
            }
            stk.push(nums[i]);
        }
        return false;
    }
}

// Accepted solution for LeetCode #456: 132 Pattern
func find132pattern(nums []int) bool {
	vk := -(1 << 30)
	stk := []int{}
	for i := len(nums) - 1; i >= 0; i-- {
		if nums[i] < vk {
			return true
		}
		for len(stk) > 0 && stk[len(stk)-1] < nums[i] {
			vk = stk[len(stk)-1]
			stk = stk[:len(stk)-1]
		}
		stk = append(stk, nums[i])
	}
	return false
}

# Accepted solution for LeetCode #456: 132 Pattern
class Solution:
    def find132pattern(self, nums: List[int]) -> bool:
        vk = -inf
        stk = []
        for x in nums[::-1]:
            if x < vk:
                return True
            while stk and stk[-1] < x:
                vk = stk.pop()
            stk.append(x)
        return False

// Accepted solution for LeetCode #456: 132 Pattern
impl Solution {
    pub fn find132pattern(nums: Vec<i32>) -> bool {
        let n = nums.len();
        let mut vk = i32::MIN;
        let mut stk = vec![];
        for i in (0..n).rev() {
            if nums[i] < vk {
                return true;
            }
            while !stk.is_empty() && stk.last().unwrap() < &nums[i] {
                vk = stk.pop().unwrap();
            }
            stk.push(nums[i]);
        }
        false
    }
}

// Accepted solution for LeetCode #456: 132 Pattern
function find132pattern(nums: number[]): boolean {
    let vk = -Infinity;
    const stk: number[] = [];
    for (let i = nums.length - 1; i >= 0; --i) {
        if (nums[i] < vk) {
            return true;
        }
        while (stk.length && stk[stk.length - 1] < nums[i]) {
            vk = stk.pop()!;
        }
        stk.push(nums[i]);
    }
    return false;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.