How do you solve 132 Pattern on LeetCode?

132 Pattern (LeetCode #456) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The 132 pattern requires careful tracking of potential candidates.

What is the brute force solution for 132 Pattern?

The brute force approach for 132 Pattern is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible triplet in the array to see if it forms a 132 pattern. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve 132 Pattern?

132 Pattern uses the following concepts: Array, Binary Search, Stack, Monotonic Stack, Ordered Set. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for 132 Pattern?

Explain your thought process clearly as you code. Consider edge cases and test your solution with them. Practice explaining your solution to someone else to solidify your understanding.

What are common mistakes when solving 132 Pattern?

Not considering the order of indices i, j, k correctly. Overlooking edge cases with negative numbers or large ranges.

#456

132 Pattern

Medium

ArrayBinary SearchStackMonotonic StackOrdered SetHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution uses a stack to keep track of potential candidates for the 132 pattern. This reduces the time complexity significantly by only needing to traverse the array once.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty stack and a variable 'second' to negative infinity.
2Step 2: Traverse the array from right to left. For each number, check if it is less than 'second'. If yes, return true (found a 132 pattern).
3Step 3: If not, while the stack is not empty and the current number is greater than the top of the stack, pop the stack and update 'second' to the popped value.
4Step 4: Push the current number onto the stack and continue until the end of the array.

solution.py10 lines

1# Full working Python code
2stack = []
3second = float('-inf')
4for num in reversed(nums):
5    if num < second:
6        return True
7    while stack and num > stack[-1]:
8        second = stack.pop()
9    stack.append(num)
10return False

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Complexity note: The time complexity is O(n) because we traverse the array once. The space complexity is O(n) due to the stack used to store elements.

1The 132 pattern requires careful tracking of potential candidates.
2Using a stack can help efficiently manage the conditions for the pattern.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.