Rotate and delete gfg potd | 02-10-24 | GFG Problem of the day

[@CodeGenius316]

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[@jvinay2210]

keep going brother it helps me a lot i watch it everyday don't stop this series plese i am new to all this

[@CS090Srikanth]

Really wonderful explanation ❤

[@ritwicksingharoy5742]

You need to perform half the times of original array size, at that time whatever the first element is is your answer. int rotateDelete(vector &arr) { // Your code here int sz=arr.size()/2; int k=1; while(n) { arr.insert(arr.begin(),arr.back()); //Last ele added to start arr.pop_back(); //removed ele from last //Array is now rotated arr.erase(arr.end()-k); //Erase k-th element from last k++; n--; } return arr[0]; }

[@omkumarjha2745]

I solved it through recursion -- TC - 0(N) and SC - 0(N) -- Recursive call stack space int solve(vector &arr , int k , int n){ if(n == 1) return 0; int ind = (k == 0) ? solve(arr , k + 1 , n) : solve(arr , k + 1 , n - 1); int deletedInd = n - k; int newInd = (ind == 0) ? n - 1 : ind; if(newInd > deletedInd){ // from example 1 return newInd; } else if(newInd == deletedInd) return newInd - 1; // from example 2 return newInd - 1; } int rotateDelete(vector &arr) { int n = arr.size(); int ind = solve(arr , 0 , n); return arr[ind]; }

[@T_M_T_K]

Subscribe kar diya sir par please java mai bhi solution samjaya karo

[@nandinigupta3836]

Keep it up!!! ❤beautiful explanation 🥹

[@manojitsaha6262]

main raat ko dekha tha tab kth element tha 2nd point pe main dry run nahi kar paya ab samjh aya ki wrong tha constrain

[@animez718]

same vro subh bhi kth element hi tha abhi change hua hoga

[@onkardhingra9083]

Yes, kth hi tha pehle

[@mohammadanas7929]

Best explain

[@OGAbhishekMalviya]

sare test case pass nhi hore

[@salmaniproductions1104]

Thank you so much

[@fshimcds]

🙏