The challenge
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.
**Follow up: **The O(n^2) is trivial, could you come up with the O(n logn) or the O(n) solution?
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.length1 <= n <= 10<sup>4</sup>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup>
The solution in Java
class Solution {
public boolean find132pattern(int[] nums) {
int n = nums.length, top = n, third = Integer.MIN_VALUE;
for (int i = n - 1; i >= 0; i--) {
if (nums[i] < third) return true;
while (top < n && nums[i] > nums[top]) third = nums[top++];
nums[--top] = nums[i];
}
return false;
}
}
You can also solve this problem by means of using a Stack.
class Solution {
public boolean find132pattern(int[] nums) {
if(nums == null || nums.length < 3){
return false;
}
int [] min = new int[nums.length];
min[0] = nums[0];
for(int i = 1; i < min.length; i++){
min[i] = Math.min(nums[i], min[i - 1]);
}
Stack<Integer> stack = new Stack<>();
for(int i = nums.length - 1; i >= 0; i--){
if(nums[i] > min[i]){
while(!stack.empty() && stack.peek() <= min[i]){
stack.pop();
}
if(!stack.empty() && stack.peek() < nums[i]){
return true;
}
stack.push(nums[i]);
}
}
return false;
}
}
