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This approach checks all possible triples (i, j, k) to determine if they form a 132 pattern. While straightforward, its complexity can be high for larger arrays due to the triple nested loop structure.
Time complexity: O(n^3) because of the three nested loops.
Space complexity: O(1) since no additional data structures are used.
1public class Solution {
2 public boolean find132pattern(int[] nums) {
3 int n = nums.length;
4 for (int i = 0; i < n - 2; i++) {
5 for (int j = i + 1; j < n - 1; j++) {
6 for (int k = j + 1; k < n; k++) {
7 if (nums[i] < nums[k] && nums[k] < nums[j]) {
8 return true;
9 }
10 }
11 }
12 }
13 return false;
14 }
15}
16This Java solution employs triple nested loops to check for a 132 pattern among elements of the array.
This approach uses a clever stack-based strategy by scanning the array from right to left. It keeps track of potential '2' and '3' candidates using a stack and a variable, optimizing the search for the '132' pattern.
Time complexity: O(n) because each element is pushed and popped once.
Space complexity: O(n) for the stack.
1defIterating over the reversed list, the stack is used to check the '132' pattern. The 'num3' variable tracks the most recent valid '2'.