Increasing Triplet Subsequence || Leetcode JAVA || O(N)

Hi folks!
Here we are solving 334. Increasing Triplet Subsequence || Leetcode C++ || O(N) time complexity || O(1) Space complexity
Easy explanation in Hindi.
Problem link : leetcode.com/p...
If you have any queries reach out to me at
linktr.ee/udya...

[@cooltomcatty]

Wow what an explanation bro! 🤩🤩🤩 God bless you 🥰🥰🥰

[@anishbishnoi29xD]

wow super bro thanks ❤️

[@udyanishere]

Thank you too

[@purna949]

Darun akta video dadavai

[@udyanishere]

Tomake aneka dhanyabada

[@purna949]

Amar poribarer aso

[@saptarshichattopadhyay8234]

@@purna949 🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣

[@ayanthisside]

great

[@udyanishere]

Thanks

[@Yugalnasare]

Thik Hai.....

[@udyanishere]

Thik Hai 👍

[@cooltomcatty]

I just found that It's similar to partion sort (card game)

[@cooltomcatty]

Wow!!!

[@rangan_dasgupta]

Dammmmmmmmmmm❤❤❤❤❤

[@udyanishere]

Lessgoo 🔥

[@asjadahmad1360]

samjhane ka tarika kaafi badhiya
keep it up bro

[@udyanishere]

Thank you so much bro.

[@vinaykaushikinvisble]

why did you check for

[@udyanojha]

According to question, We want to return true only if nums[i] < nums[j] < nums[k]

[@udyanojha]

and by using

[@udyanojha]

Dry run for 1 2 1 2 1 2 3 case

[@paras0verma]

😎😎

[@udyanishere]

// Python3 :
class Solution:
def increasingTriplet(self, nums: List[int]) -> bool:
left, mid = sys.maxsize, sys.maxsize
for num in nums:
if num

[@udyanishere]

// C++ :
class Solution {
public:
bool increasingTriplet(vector& nums) {
int N = nums.size();
int left = INT_MAX;
int mid = INT_MAX;

for(int num : nums) {
// num = nums[i];
if(num

[@udyanishere]

// JAVA :
class Solution {
public boolean increasingTriplet(int[] nums) {
int N = nums.length;
int left = Integer.MAX_VALUE;
int mid = Integer.MAX_VALUE;

for(int num : nums) {
// num = nums[i];
if(num