# 334. Increasing Triplet Subsequence
Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.
Formally the function should:
Return true if there exists i, j, k
such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.
Note: Your algorithm should run in O(n) time complexity and O(1) space complexity.
Example 1:
Input: [1,2,3,4,5]
Output: true
Example 2:
Input: [5,4,3,2,1]
Output: false
# Solution
Approach 1: O(nlogn).
Approach 2: O(n) by tracking the mininum and the second mininum elements.
# Code (Python)
Approach 1:
Approach 2:
# Code (C++)
Approach 1:
// O(nlogn) time.
class Solution {
public:
bool increasingTriplet(vector<int>& nums) {
vector<int> buf;
for (int i = 0; i < nums.size(); ++i)
{
int lo = 0;
int hi = buf.size() - 1;
while (lo <= hi)
{
int mid = lo + (hi - lo) / 2;
if (nums[i] < buf[mid])
hi = mid - 1;
else if (nums[i] > buf[mid])
lo = mid + 1;
else
break;
}
if (lo > (int)buf.size() - 1)
buf.push_back(nums[i]);
else if (lo > hi)
buf[lo] = nums[i];
}
return buf.size() >= 3;
}
};
Approach 2:
// O(n) time.
class Solution {
public:
bool increasingTriplet(vector<int>& nums) {
int min1Idx = 0;
int min2Idx = -1;
for (int i = 1; i < nums.size(); ++i)
{
if (min2Idx >= 0 && nums[i] > nums[min2Idx])
return true;
// else if (nums[i] <= nums[min1Idx])
// min1Idx = i;
// else if (min2Idx < 0 || nums[i] < nums[min2Idx])
// min2Idx = i;
else if (nums[i] > nums[min1Idx])
min2Idx = i;
else if (nums[i] < nums[min1Idx])
min1Idx = i;
}
return false;
}
};