132 Pattern

class Solution {
public:
    bool check132pattern(vector<int>& elements) {
        int n = elements.size();
        if (n < 3) {
            return false;
        }
        vector<int> leftMin(n);
        leftMin[0] = elements[0];

        for (int i = 1; i < n; i++) {
            leftMin[i] = min(leftMin[i - 1], elements[i]);
        }

        for (int j = n - 1, k = n; j > 0; j--) {
            if (elements[j] <= leftMin[j]) continue;
            while (k < n && elements[k] <= leftMin[j]) k++;
            if (k < n && elements[k] < elements[j]) {
                return true;
            }
            elements[--k] = elements[j];
        }
        return false;
    }
};

The given C++ code defines a method check132pattern within a class named Solution. This method aims to determine if there exists a subsequence of at least three numbers in the given array elements that can form a pattern of 132. Here's how the solution works:

• The method starts by determining the size of the input vector elements. If the size is less than three, it immediately returns false as no triplet can exist.
• An auxiliary vector leftMin is created to keep track of the minimum values from the left up to the current position. The first element is initialized to the first element of elements.
• A loop fills the leftMin vector where each position holds the minimum value found so far from the start of the array to that position.
• Another loop starts from the end of the array and uses variables j and k to check for the existence of the 132 pattern. The variable j traverses the array from the last but one element to the start, and k is used as a reference to scan for a suitable third element of the pattern.
  • The loop checks if the current element elements[j] is greater than the smallest element to its left leftMin[j]. If not, it skips the current iteration.
  • A nested while loop increments k to find an element larger than the leftMin[j] but less than elements[j].
  • If an element fulfilling the 132 condition is found (elements[k] < elements[j]), the function returns true.
• If the loops complete without finding a suitable triplet, the method returns false.

Thus, the function efficiently checks for the 132 pattern in the array by minimizing redundant checks and optimizing element comparisons with the help of the leftMin vector and careful placement of the iterators j and k.

public class Solution {
    public boolean checkPattern(int[] arr) {
        if (arr.length < 3) return false;
        int[] minValues = new int[arr.length];
        minValues[0] = arr[0];
        for (int i = 1; i < arr.length; i++) {
            minValues[i] = Math.min(minValues[i - 1], arr[i]);
        }
        for (int j = arr.length - 1, k = arr.length; j >= 0; j--) {
            if (arr[j] > minValues[j]) {
                while (k < arr.length && arr[k] <= minValues[j]) k++;
                if (k < arr.length && arr[k] < arr[j]) return true;
                arr[--k] = arr[j];
            }
        }
        return false;
    }
}

In this Java solution, you will check for the presence of a 132 pattern in a given array of integers. The 132 pattern is identified where an element n1 is followed by n2, which is followed by n3, and n1 < n3 < n2. This can be approached systematically:

• First, verify if the array contains at least three elements to form a valid pattern. If not, immediately return false.
• Create an array minValues to store the minimum value found up to each index of the original array. Initialize the first element of minValues with the first element of the input array.
• Populate the minValues array by maintaining the minimum value seen so far as you iterate through the array.
• Then, traverse the array from the last element towards the first. Use two pointers: j for the current position being checked in the original array and k initialized beyond the end of the array.
• For each element, compare the current element with the minimum value at that index. If the current element is greater:
  • Check if there exists a suitable n3 (middle element in the 132 pattern). Move k such that it points to the potential n3 while ensuring n3 remains greater than the minimum value at j.
  • If a valid n3 is found that is also less than the current element arr[j], immediately return true.
  • If not found, update the index k to the current index j, effectively moving backwards to check for other potential n3 values.
• If no valid pattern is detected through the entire array, conclude by returning false.

This algorithm smartly checks the conditions for the 132 pattern using a backward iteration combined with a dynamic update of minimum values and a controlled search for the middle pattern element, leveraging both space and time efficiencies inherent in the process. Ensure to understand the conditions correctly and implement careful index management when adapting or expanding this solution.

class Solution:
    def findPattern(self, values: List[int]) -> bool:
        if len(values) < 3:
            return False
        lowest = [-1] * len(values)
        lowest[0] = values[0]
        for index in range(1, len(values)):
            lowest[index] = min(lowest[index - 1], values[index])

        right = len(values)
        for middle in range(len(values) - 1, -1, -1):
            if values[middle] <= lowest[middle]:
                continue
            while right < len(values) and values[right] <= lowest[middle]:
                right += 1
            if right < len(values) and values[right] < values[middle]:
                return True
            right -= 1
            values[right] = values[middle]
        return False

The provided Python solution efficiently determines if there exists a 132 pattern in a given list of integers. This pattern exists if there is a subsequence of three integers i, j, and k such that i < k < j. The implementation encapsulates the logic within a class named Solution using the method findPattern.

Here's how the algorithm operates:

1. First, check if the input list contains fewer than three elements. If true, immediately return False as it's impossible to find a valid subsequence with less than three numbers.
2. Initialize an auxiliary list lowest[] that stores the minimum value encountered up to each index of the array.
3. Iterate through the values to populate the lowest array while comparing each element and maintaining the lowest value found up to the current position.
4. Use a stack-like approach with a pointer right set to the end of the list and iterate backward through the list, which allows efficient checking against the conditions of the 132 pattern.
5. For each "middle" element in the reversed loop:
   • Skip over elements that do not satisfy the 132 middle value condition (if they are equal or less than their corresponding lowest value).
   • While maintaining the desired order, adjust the right pointer and update values to find a suitable "k".
   • If conditions for i, j, and k are met, return True.
6. If no patterns are found across all iterations, return False.

The dual scan approach—forward to create the lowest array and backward to check for pattern conditions—ensures that the method is efficient. Handling elements in this way exploits the nested nature of the pattern, streamlining detection by minimizing unnecessary comparisons.