Key Pair Hashing Problems |

[@xyz956]

op admi ho bhai sahi mein,bhai hindi me lkin samjhane ke baad alag hi feel aata hai.

[@HelloWorldbyprince]

Exactly 😂😌

[@vinayjoshi8714]

Bhai question padha aur ho gaya Bina dekhe solution aapke padhane ki wajah se hi kar paya mein Thank you bhaiya ❤

[@nayansinghal23]

Bhaiya Hashing was not taught to me by my university professor and when I opened gfg i was unable to understand hashing. But after this playlist I am becoming confident day by day and can solve easy problems of hashing from gfg. All credit goes to you. Thank you bhaiya🥰🥰🥰

[@HelloWorldbyprince]

Thanks a lot Nayan Please, if possible then share this channel in your college, groups, or LinkedIn bcoz it's cost nothing to you 😀

[@nayansinghal23]

@@HelloWorldbyprince bhaiya aapke channel ke bare m sabko pata h junior Senior everyone. ❣️❣️

[@oqant0424]

after seeing this playlist I am becoming confident day by day and can solve problems of hashing .All credit goes to you. Thank you bhai!!!!!!!!!!!!!!!!!

[@bhupeshpattanaik7150]

Congratulations for the sponsorship 🥳🔥

[@PriyankaJaiswal-gf1rw]

awesome explaination bhaiya. I was unable to do one of the hashing question but after watching this video i did it in no time.tysm bhaiya

[@HelloWorldbyprince]

Keep learning priyanka After this playlist of hashing You feel more confident 😃

[@PriyankaJaiswal-gf1rw]

Yes bhaiya.Tysm

[@Man_of_Culture.]

Hashing is good but it's solution with sorting is best .

[@shubhamgupta9778]

bool hasArrayTwoCandidates(int arr[], int n, int sum) { unordered_set s; for (int i = 0; i < n ; i++){ int temp = sum - arr[i]; if (s.find(temp) != s.end()) return true; s.insert(arr[i]); } return false; // code here } we can solve this question by using unordered_set also bro...

[@PIYUSH-lz1zq]

Bro >1 ka conditions kaha hai ...! If sum =8 and array might have 4 as element then it will count twice of that element !!

[@mithileshkumarsingh222]

USING UNORDERED SET unordered_set st; for(int i=0; i

[@HelloWorldbyprince]

Nice buddy 🔥

[@cr7johnChan]

Thanl you bhaiya for guiding !

[@HelloWorldbyprince]

Thanks for the supporting me

[@kritisingh3194]

Java solution for the question: boolean hasArrayTwoCandidates(int arr[], int n, int X) { Map mp = new HashMap(); for(int i=0; i

[@HelloWorldbyprince]

amazing work Kriti

[@masumali8356]

you are the best.............masum.

[@ayushkashyap3643]

alternate solution- unordered_setstore; store.insert(arr[0]); for(int i=1;i

[@AbuBakar-tx7wj]

It can be done by unordered_set bool hasArrayTwoCandidates(int arr[], int n, int x) { unordered_set s; for(int i=0;i

[@kumarabhishek3890]

we can use a hash set too just temporarily remove that element before assessing it.

[@oqant0424]

thanks bhaiya for always inspiring us :)

[@HelloWorldbyprince]

Always welcome

[@nagabasavannas1440]

your videos are best in youtube and explation is better than our profesors. by the way can't we solve the problem using unordered_set when hit with element erasing it ??????

[@HelloWorldbyprince]

Thanks for the support ☺️ And yes, if u think u can just give a try and if it is possible then yess we can do it And please also paste the solution in the comments section

[@ovokteb9671]

Two pointer approach would also be a optimal solution for this problem:) And thank u bhaiya ❤ for this awesome playlist on Hashing 🙌

[@altafmazhar7762]

No it wont to use two pointer we first have to sort the array

[@HelloWorldbyprince]

Most welcome 😊

[@HelloWorldbyprince]

u r right

[@only_for_fun1234r]

using set ----------- bool hasArrayTwoCandidates(int arr[], int n, int x) { set st; for(int i=0;i

[@uditpanjiyar3203]

But we can also solve this question using unordered set & here is the simplest code bool hasArrayTwoCandidates(int arr[], int n, int x) { // code here unordered_sets; for(int i=0; i

[@shashwatdev2371]

It can also be done using Unordered set class Solution{ public: // Function to check if array has 2 elements // whose sum is equal to the given value bool hasArrayTwoCandidates(int arr[], int n, int x) { unordered_setuset; int j,k; for(int i=0;i

[@HelloWorldbyprince]

yup

[@dudlavarumakanth4671]

super prince sir

[@HelloWorldbyprince]

Welcome 😃😃

[@NIKHILSINGH-kg2du]

good explanation

[@HelloWorldbyprince]

Thanks a lot nikhil

[@deepanshu4769]

alternate solution bool hasArrayTwoCandidates(int arr[], int n, int target) { unordered_map umap; for(int i=0;i

[@princesharma7623]

SIMPLE SOLUTION unordered_mapm; for(int i=0; i

[@aryangupta7290]

Alternatively start graph as well, sir.

[@HelloWorldbyprince]

Yeah buddy I will try

[@shivamsinha5554]

my approach hope this will help you

[@amitprasadiitg2564]

CPP unordered_mapm; for(int i=0;i

[@HelloWorldbyprince]

nice bro ..keep learning like this

[@pratik.784]

Set se easily ho jayega

[@oqant0424]

11/18 done (8.12.22)

[@AinasDiaries]

Isn't it two sum problem?

[@HelloWorldbyprince]

Yup

[@madhukareppalapelly7253]

yes we can solve this using two pointer technique also

[@sagarsingh-uh6xd]

unordered_map hash; for (int i = 0; i < n; i++) { int numberToFind = x - arr[i]; //if numberToFind is found in map, return them if (hash.find(numberToFind) != hash.end()) { return true; } //number was not found. Put it in the map. hash[arr[i]] = i; } return false; plz explain this . second comment ??

[@HelloWorldbyprince]

Wahh ekdum sahi 😊

[@MilindGupta]

Is it two sum problem of leetcodr?

[@HelloWorldbyprince]

Yess

[@maniish-sgh]

Done !!

[@surgeonofdeath8199]

// whose sum is equal to the given value bool hasArrayTwoCandidates(int arr[], int n, int x) { // code here unordered_set mpp; int flag=0; for(int i=0;i

[@kayumkhan_kks]

10:5 hum index bhi check kar skthe hai

[@Ankur-zd4db]

Bhaiya java mai bhi bta diya karo.

[@anujpradhan480]

Sirf if size of the arr is 1???

[@HelloWorldbyprince]

Then pair kaise banega ? ye puchna chah rhe ho app

[@anujpradhan480]

@@HelloWorldbyprince ok sir.

[@316_mohammadahadkhan7]

unordered_map s; for (int i=0;i

[@prakhar266]

done

[@HelloWorldbyprince]

good

[@masumali8356]

sort(arr,arr+n); int i=0; int j=n-1; while(i

[@ishandas2674]

you have a very good video but you talk unnecessary things in between which breaks the concentration. kindly work on it otherwise is fire!

[@HelloWorldbyprince]

Will try to do better in future videos 🙏

[@oqant0424]

if u are someone jisko ye sab ni samjh aa raha.............bhai plz jake puri playlist khatam karo.....sawal makkhan ki tarah solve hone lagenge ...i can bet u!!!!!!!!!!!!! thank me later :)

[@ayushdhiman8329]

yai thik hai ya nhi class Solution { boolean hasArrayTwoCandidates(int arr[], int x) { // code here HashSet set = new HashSet(); for(int i=0; i

[@tusharkarle7870]

Bhai Insta Dekh lo please