LeetCode in Kotlin

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

Solution

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

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.
     */
    fun find132pattern(nums: IntArray): Boolean {
        val stack: Deque<Int> = LinkedList()
        var s3 = Int.MIN_VALUE
        for (i in nums.indices.reversed()) {
            if (nums[i] < s3) {
                return true
            } else {
                while (stack.isNotEmpty() && nums[i] > stack.peek()) {
                    s3 = Math.max(s3, stack.pop())
                }
            }
            stack.push(nums[i])
        }
        return false
    }
}

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