Perfect Squares | LeetCode 279 | C++, Java, Python

[@srinish1993]

11:26 integration explanation was 👌

[@KnowledgeCenter]

Thanks.

[@crisag.2698]

Thank you so much for this video. What would be the brute-force recursive time complexity?

[@paragroy5359]

Nice explanation sir.....thanks for the video

[@KnowledgeCenter]

Glad you liked it

[@siddharthsingh4330]

what is 1+dp[x-sq] doing I didn't understand properly?

[@raviashwin1157]

Respected sir,which is better platform to code in leetcode or codeforces,will it be making any difference in placement if i just code in leetcode but not in codeforces.Is codeforces rating needs to be too good to grab good placement?

[@midhileshmomidi2434]

Codeforces for Competitive programming and leetcode for interviews I felt codeforces problems are tougher than leetcode

[@KnowledgeCenter]

Codforces ratings are not related to placements. For placements, LeetCode questions are more relevant.

[@raviashwin1157]

@@KnowledgeCenter Sir can you please make a video how to use leetcode effectively

[@mrlectus]

Any brute force solution?

[@mohammedshoaib1635]

Great explanation!

[@KnowledgeCenter]

Thanks!

[@ijaz2020]

Reconstruct Itinerary waiting for this video. I checked old videos now it is not there as well.

[@KnowledgeCenter]

Have cramp in Neck and back. Was sleeping. Will try to add it before sleeping.

[@mukultaneja7243]

I tried it's memoization, and it's faster than the tabulation. Please explain why...????

[@sehajverma6952]

Just interpret this question as Coin change problem with infinite number of same denomination coins. class Solution { public: int numSquares(int n) { int x = sqrt(n); int dp[x+1][n+1]; memset(dp,0,sizeof(dp)); int m=INT_MAX; for(int i=1;i

[@KnowledgeCenter]

Exactly.

[@shubhakardebroy2111]

According to Lagrange's four square theorem we can solve this in O(N) www.geeksforgeeks.org/lagranges-four-square-theorem/ Solution in python: def numSquares(self, n): sqr = sqrt(n) pool = {i**2 for i in range(int(sqr)+1)} test = [i**2 for i in range(int(sqr*0.71)+1)] for i in test: for j in test: if n-i-j in pool: return 3-(i==0)-(j==0) return 4

[@KnowledgeCenter]

Thanks for sharing.