just found a gold mine ,thanks for the content............ Bro i just wanted to ask most of the time ,most of the time i am able to figure out the logic but implementation of it is where i get i get stuck...... how can i overcome this

nice explanation but can you make it a bit more interactive for beginners(i was so sleepy after watching this lecture).

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Nice bro 👏Explained perfectly , easy and understandable.

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plss provide the name of drawing extension that you use

sir the space complexity will be o(n) but in question space should beO(1)

Easy peasy Q's, i solved this using sliding window-O(n,1).

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noice❤

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Nice Explanation❣️, it's getting so well day by day 👏🏻

Thanks for the solution brother!! Very well explained!!😊😊😊😊

W

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Please daily vedio laiyega

Thanks for your guidance. Please upload playlist on Advance DSA with Java. Topics like Tree,Graph,etc.

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class Solution: def pattern(self, N): # code here def helper(n, flag): if flag: if n <= 0: return helper(n, False) else: return [n] + helper(n - 5, True) else: if n > N: return [] else: return [n] + helper(n + 5, False) return helper(N, True) Bro I stucked here from 2 hours. It is gfg recursion, problem name is print pattern. My code time limit exceeded. Please help me bro. Please give optimized version

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Please provide the code for second approach..........

English please, you are having audience from south also....

NICE SUPER EXCELLENT MOTIVATED

Plz continue in english iam from tamil nadu

teach in hindi

very nice but sir pls continue with hindi cause approach hindi m hi connect ho pata hai

please provide the code in python🐍 also in your github class Solution: def countTriplets(self, arr, target): n = len(arr) ans = 0 for i in range(n - 2): j, k = i + 1, n - 1 while j < k: total = arr[i] + arr[j] + arr[k] if total < target: j += 1 elif total > target: k -= 1 else: e1, e2 = arr[j], arr[k] c1, c2 = 0, 0 # Count occurrences of e1 while j <= k and arr[j] == e1: c1 += 1 j += 1 # Count occurrences of e2 while j <= k and arr[k] == e2: c2 += 1 k -= 1 if e1 == e2: ans += (c1 * (c1 - 1)) // 2 else: ans += c1 * c2 return ans

sir please give the link of that first hashmap explanation

explain in english..

Nice explanation 💖💖, But can we get next videos in little louder voice. 😊

Also one request, try to explain everyday's solution with unordered map as well if it is applicable, because that way people will get familiar with it. I see that the last three day's problems can be simply solved with unordered_map including today's So I would suggest, you can explain the approach which you wanted, and add a bonus part of the video where you can explain the unordered_map approach. I saw the below code in GfG comments section int countTriplets(vector<int>& arr, int target) { unordered_map<int, int> freq; for (int e : arr) { freq[e]++; } int ans = 0; for (int i = 0; i < arr.size(); i++) { freq[arr[i]]--; // Remove the current element from the frequency map for (int j = 0; j < i; j++) { // Traverse from 0 to i (exclusive) int lookfor = target - arr[i] - arr[j]; if (freq.find(lookfor) != freq.end()) { ans += freq[lookfor]; } } } return ans; } But it is difficult to follow and understand for a beginner, still the solution seems more simpler Thanks for your efforts

Please continue this series it really help student like us.

Hi, Great Video, Thanks, And please continue teaching in English

potd in gfg ek fix time ke baad(time exceed) usko solve karne pr usko consider nahi kiya jata hai kya???? i solved the problem but more than the time limit so they didn't give me geekbits and also not continue my streak

Teach only in english sir I m from tamilnadu if uh teach in hindi i don't get the things Moreover I watch uh're video only i do daily challenge

contrain maximum which is applied is till 10^5 in the question ,so how can we choose this 10^6 constraint using two for loops ?

Initially I used binary search but because of duplicate elements it gave me WA...

Thanks Bro for explaining

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nice