• Approach 1: Brute Force
• Approach 2: Using Stacks

The brute force approach involves checking every possible triplet to see if it forms a 132 pattern. The 132 pattern means finding three indices i, j, k such that i < j < k and nums[i] < nums[k] < nums[j].

1. Loop over the array and fix i.
2. For every fixed i, fix j and ensure j > i.
3. For every fixed j, loop over the array to find an element k, where k > j and nums[i] < nums[k] < nums[j].
4. If such a triplet is found, return true.
5. If after checking all possibilities, no triplet is found, return false.

public boolean find132pattern(int[] nums) {
    for (int i = 0; i < nums.length; i++) {
        for (int j = i + 1; j < nums.length; j++) {
            for (int k = j + 1; k < nums.length; k++) {
                // Check the pattern condition
                if (nums[i] < nums[k] && nums[k] < nums[j]) {
                    return true;
                }
            }
        }
    }
    return false;
}

• Time Complexity: O(n^3), where n is the length of the array, due to the three nested loops.
• Space Complexity: O(1), no extra space used.

To optimize, we can use a stack to maintain candidates for the second largest number (nums[j]) in the pattern 132 by iterating from the right. We will also maintain a variable nums[k] as the second number in the 132 pattern.

1. Traverse the array from the right.
2. Keep a variable third to track the maximum number less than nums[j] which can be a candidate for nums[k].
3. Use a stack to maintain the numbers which can potentially be nums[j].
4. For each element nums[i], check:
   • If nums[i] is less than third, a valid 132 pattern is found.
5. If no 132 pattern is found by the end of the loop, return false.

public boolean find132pattern(int[] nums) {
    if (nums.length < 3) return false;
    
    int third = Integer.MIN_VALUE;
    Stack<Integer> stack = new Stack<>();
    
    for (int i = nums.length - 1; i >= 0; i--) {
        // Check if the current number can be nums[i] in 132 pattern
        if (nums[i] < third) {
            return true;
        }
        
        // Maintain the stack such that any number in it is a potential nums[j]
        // And update third to be the maximum valid nums[k]
        while (!stack.isEmpty() && nums[i] > stack.peek()) {
            third = stack.pop();
        }
        
        // Push the current number as potential nums[j]
        stack.push(nums[i]);
    }
    
    return false;
}

• Time Complexity: O(n), since each element is pushed and popped from the stack at most once.
• Space Complexity: O(n), for the stack used to store potential nums[j] candidates.