LeetCode 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.
Your algorithm should run in O(n) time complexity and O(1) space complexity.

Examples:

Given

[1, 2, 3, 4, 5]

,

return

true

.

Given

[5, 4, 3, 2, 1]

,

return

false

.

#include <vector>
#include <iostream>
#include <climits>
using namespace std;

// increasing triplet subsequence.
/*
Given an unsorted array return whether an increasing subsequence of length 3
exists or not in the array.
return true if there exists i, j, k
such that arr[i] < arr[j] < arr[k] given that 0 < i < j < k <= n-1
else return false.
*/
// requirment: O(n) time and O(1) space.
// if we do the same way as longest increasing subsequence.
// this method will take O(n) space and O(N2) time complexity.
bool increasingTriplet(vector<int>& nums) {
if(nums.size() < 3) return false;
int n = nums.size();
vector<int> dp(n+1, 1);
for(int i = 1; i <= n; ++i) {
for(int j = 0; j < i; ++j) {
if(nums[i] > nums[j]) {
dp[i] = max(dp[i], dp[j] + 1);
}
}
}
for(int i = 0; i <= n; ++i) {
if(dp[i] >= 3) return true;
}
return false;
}

bool increasingTripletII(vector<int>& nums) {
if(nums.size() < 3) return false;
int min = INT_MAX;
int mid = INT_MAX;
for(auto n : nums) {
if(n < min) {
min = n;
} else if(n > min) {
if(mid < n) return true;
mid = n;
}
}
return false;
}

int main(void) {
vector<int> nums{3, 4, 0, 2, 1, 5};
cout << increasingTripletII(nums) << endl;
}