How to Solve the 132 Pattern in Java

The challenge

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.

Follow up: The O(n^2) is trivial, could you come up with the O(n logn) or the O(n) solution?

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 <= 104
  • -109 <= nums[i] <= 109

The solution in Java

class Solution { public boolean find132pattern(int[] nums) { int n = nums.length, top = n, third = Integer.MIN_VALUE; for (int i = n - 1; i >= 0; i--) { if (nums[i] < third) return true; while (top < n && nums[i] > nums[top]) third = nums[top++]; nums[--top] = nums[i]; } return false; } }
Code language: Java (java)

You can also solve this problem by means of using a Stack.

class Solution { public boolean find132pattern(int[] nums) { if(nums == null || nums.length < 3){ return false; } int [] min = new int[nums.length]; min[0] = nums[0]; for(int i = 1; i < min.length; i++){ min[i] = Math.min(nums[i], min[i - 1]); } Stack<Integer> stack = new Stack<>(); for(int i = nums.length - 1; i >= 0; i--){ if(nums[i] > min[i]){ while(!stack.empty() && stack.peek() <= min[i]){ stack.pop(); } if(!stack.empty() && stack.peek() < nums[i]){ return true; } stack.push(nums[i]); } } return false; } }
Code language: Java (java)
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