_456.java

package com.fishercoder.solutions;

import java.util.Stack;
/**Given a sequence of n integers a1, a2, ..., an, a 132 pattern is a subsequence ai, aj, ak such that i < j < k and ai < ak < aj. Design an algorithm that takes a list of n numbers as input and checks whether there is a 132 pattern in the list.

 Note: n will be less than 15,000.

 Example 1:
 Input: [1, 2, 3, 4]

 Output: False

 Explanation: There is no 132 pattern in the sequence.
 Example 2:
 Input: [3, 1, 4, 2]

 Output: True

 Explanation: There is a 132 pattern in the sequence: [1, 4, 2].
 Example 3:
 Input: [-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].*/
public class _456 {

    /**Looked at this post: https://discuss.leetcode.com/topic/67881/single-pass-c-o-n-space-and-time-solution-8-lines-with-detailed-explanation
     * It scans only once, this is the power of using correct data structure.
     * It goes from the right to the left. It keeps pushing elements into the stack, but it also keeps poping elements out of the stack as long as the current element is bigger than this number.*/
    public static boolean find132pattern(int[] nums) {
        Stack<Integer> stack = new Stack();
        
        int s3 = Integer.MIN_VALUE;
        for (int i = nums.length-1; i >= 0; i--){
            if (nums[i] < s3) return true;
            else while (!stack.isEmpty() && nums[i] > stack.peek()){
                s3 = Math.max(s3, stack.pop());
            }
            stack.push(nums[i]);
        }
        
        return false;
    }

    public static void main (String...args){
        int[] nums = new int[]{-1, 3, 2, 0};
        System.out.println(find132pattern(nums));
    }
}