Two Pointer Technique | Competitive Programming

Hey Guys, In this video I have explained two pointers technique which is helpful in Competitive Programming and coding interviews.
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[@ashishcode]

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[@bebo5619]

Thank you very much, I am on a time crunch studying for the second round of a huge interview and your video made the concept very simple and understandable. You're amazing, keep doing what you are doing!

[@GamingUnitesGR]

how did the interview go?

[@Asli_Investing]

Nicely Explained :) Saved me some time by building the intuition quickly

[@gouravm4986]

Please make a playlist of these things.. Would be really helpful.. I solved two question myself after watching this video so kinda very helpful for me.. Also shared your video to my friends. 😊thank you sir

[@ashishcode]

Thank you so much for your kind words.. I will be sure to continue!

[@tentx652]

hey bro ....the way u explain is simply awesome.Covering all the edge points is great ,ATB man!!

[@hyperoid3901]

Thank you very much Ashish, well explained with your demonstration.

[@sandeepaksingh9248]

Very nice. Keep it up make a series of competitive programming.

[@bhushanmendhe7904]

Loved the way you explained! Please bring more such videos on more such frequently used algorithms :)

[@ashishcode]

Thank youu, sure I will.

[@JackCoderr]

Thank you so much bhaiya...you are a great teacher ❤❤❤

[@saurabhnikam]

Really helpful .....good explanation.

[@nileshchauhan7117]

Thanku so uch ashihs plz mak this more and more

[@saransappa603]

Nice explanation Asish Kumar.

[@dowenrobinson1589]

Excellent explanation

[@akshaysb526]

great explanation

[@shreelbansal3026]

Nice!!

[@PrashantSingh-pg9vq]

what would be your explaination for this case? [0,2,4,5] diff=1;

[@A5IR]

i had the same thought. for this case, (2 - 0) will be 2, which is greater than 1. so we would increment the left pointer and now both pointers are at the same index. so I think we need to check if (i == j) we increment j. the while loop keeps going while j is less than length of array.

[@Noone-kl6sc]

bool findPair(int arr[], int size, int n){ sort(arr, arr+size); int i=0; int j=1; while(i

[@midhileshmomidi2434]

When the array is not sorted then we have to use hashmap??

[@ashishcode]

You can either use a hasmap or sort the array and then use two pointers technique.

[@selvaragavan_10]

will two difference works for this test case? [1,5,9,13,14] target = 1;

[@manishsharma2211]

Good One, thanks

[@rohitpol]

Nice explanation bro 😎

[@omkarpasalkar47]

Thank u so much brother

[@tusharpandey6584]

very nice explanation thx :3

[@rba1aji216]

can i use this two pointer technique in unsorted array??

[@ashishcode]

No, it works only on sorted arrays.

[@fuehrercheem]

Couldn't get better

[@udbhavvikramsingh3449]

how it is O(nlogn) time complexity someone explain plz why not (n*n)

[@ashishcode]

It's O(nlogn) because we use sorting and sorting takes nlogn and then we have used only one traversal meaning only one for loop which takes O(n) so it becomes totally O(nlogn)

[@udbhavvikramsingh3449]

@@ashishcode got it

[@saibharath780]

In hash map tech you used it contains liner search so it might take more time know

[@saibharath780]

Bro can you give reply I am right or wrong

[@ashishcode]

No,using hashmap it takes O(n) that is linear time which is better than sorting and then using two pointers which takes O(nlogn). If it's already sorted it will take O(n) which is same for using hashmap but hashmap will give O(n) even if not sorted. Sorry for the late reply.

[@RN-jo8zt]

Thanks

[@Python_DSA_Journey]

please tell the same with hashing

[@joelcruz2873]

With a Map, you could place the index and value of each array element inside the Map. Then for each element, you subtract it from the target and look for that difference inside of the map (which will be O(1)). In the worst case you'll have to do this O(N) times.