g0401_0500.s0456_132_pattern.Solution Maven / Gradle / Ivy

Go to download

Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation

Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.38

package g0401_0500.s0456_132_pattern;

// #Medium #Array #Binary_Search #Stack #Ordered_Set #Monotonic_Stack #Udemy_Arrays
// #2022_07_18_Time_16_ms_(82.41%)_Space_71.9_MB_(8.42%)

import java.util.Deque;
import java.util.LinkedList;

/**
 * 456 - 132 Pattern\.
 *
 * Medium
 *
 * 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
**/
public class Solution {
    /*
     * 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 boolean find132pattern(int[] nums) {
        Deque stack = new LinkedList<>();
        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;
    }
}