456. 132 Pattern
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/**
* Time complexity : O(N^2). Two loops are used to find the nums[j], nums[k]
* pairs. Here, n refers to the size of nums array.
* Space complexity : O(1)
*/
class Solution {
public boolean find132pattern(int[] nums) {
if(nums == null || nums.length < 3) {
return false;
}
int n = nums.length;
int min = Integer.MAX_VALUE;
for(int j = 0; j < n-1; j++) {
min = Math.min(min, nums[j]);
for(int k = j+1; k < n; k++) {
if(min < nums[k] && nums[k] < nums[j]) {
return true;
}
}
}
return false;
}
}