132 Pattern
Problem Description
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: trueExplanation: 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
Solve it on LeetCode
Approaches
1. Brute Force
Intuition:
The brute force approach involves checking every possible triplet to see if it forms a 132 pattern. The 132 pattern means finding three indices i, j, k such that i < j < k and nums[i] < nums[k] < nums[j].
Steps:
- Loop over the array and fix
i. - For every fixed
i, fixjand ensurej > i. - For every fixed
j, loop over the array to find an elementk, wherek > jandnums[i] < nums[k] < nums[j]. - If such a triplet is found, return true.
- If after checking all possibilities, no triplet is found, return false.
Code:
- Time Complexity: O(n^3), where
nis the length of the array, due to the three nested loops. - Space Complexity: O(1), no extra space used.
2. Using Stacks
Intuition:
To optimize, we can use a stack to maintain candidates for the second largest number (nums[j]) in the pattern 132 by iterating from the right. We will also maintain a variable nums[k] as the second number in the 132 pattern.
Steps:
- Traverse the array from the right.
- Keep a variable
thirdto track the maximum number less thannums[j]which can be a candidate fornums[k]. - Use a stack to maintain the numbers which can potentially be
nums[j]. - For each element
nums[i], check: - If
nums[i]is less thanthird, a valid 132 pattern is found. - If no 132 pattern is found by the end of the loop, return false.
Code:
- Time Complexity: O(n), since each element is pushed and popped from the stack at most once.
- Space Complexity: O(n), for the stack used to store potential
nums[j]candidates.