Trilogy Intern Hunt Online Round Discussion (18th Dec. '22)

I couldn't find any discussion or editorial blog, so am creating a new one.
Was able to solve the 2nd problem with Segment Tree (standard stuff) and attempted to solve the first one with 6D DP (wasn't able to debug an RE).
It would be great if you could share your approaches for the four problems.

• 24 hours ago, # ^ | No, I don't have them. Thought they'd post an editorial or something. The IDE was kind of unintuitive as well, had to rewrite more than 60 lines of code because the text-area wouldn't accept any text that I had written in my editor. Plus there was a grey area around whether one could create additional functions and classes, or only the solution function could be modified. The testing apparatus could have also been better. Overall, not a great experience.

  • 23 hours ago, # ^ | I also hate the IDE of InterviewBit. Worst experience I have with the IDE.

    • 20 hours ago, # ^ | Exactly its soo hard to debug there!

• 23 hours ago, # ^ | Could someone share the intended approach / an elegant solution? Mine was clumsy and complicated and full of bugs; 6D DP's never a good idea, ig.

  • 23 hours ago, # ^ | Mine was 5D digit dp and cancerous as it involved finding the maximum and least set bit of a number. The 6d digit dp is better and easier to write.

    • 23 hours ago, # ^ | Did u get ac for this?

      • 23 hours ago, # ^ |

        • 23 hours ago, # ^ | can you share the solution of 6-D dp?

        • 18 hours ago, # ^ | do you have any idea what will be the final cutoff for the shortlisting process?

    • 23 hours ago, # ^ | Could you please share the DP approach?

  • 23 hours ago, # ^ | ← Rev. 3 →   +1 Mine is 6D Dp. Could not find the bug in contest time Later found it.

  • 22 hours ago, # ^ | ← Rev. 2 →   0 I dont know what is wrong with my code, but I have used 5-D dp and used digit DP . Is there any silly mistake? or some wrong approach? int dp[18][10][10][16][2]; const int N = 1e9+7; int get(int pos , string &s, int mx, int mn, int xr, int tgt){ if(pos == 0) return (2*xr>mx+mn); if(dp[pos][mx][mn][xr][tgt] != 1) return dp[pos][mx][mn][xr][tgt]; int ans = 0, ub = (tgt? (s[s.size()-pos]-'0'):9); for(int i = 0; i <= ub; i++){ ans += get(pos-1,s,max(mx,i),min(mn,i),(xr^i),tgt&(i==ub)); ans %=N; } return dp[pos][mx][mn][xr][tgt] = ans; } int Solution::solve(string A, string B){ memset(dp,-1,sizeof(dp)); int a = get(A.size(),A,0,9,0,1); memset(dp,-1,sizeof(dp)); int b = get(B.size(),B,0,9,0,1); int mx=0,mn=9,xr=0; for(auto it: A)mx=max(mx,it-'0'),mn=min(mn,it-'0'),xr^=(it-'0'); if(2*xr>mx+mn)ans--; return ans; }

    • 22 hours ago, # ^ | You need one more state known to handle cases where you haven't started building numbers. For example if you start building from the second digit and put a 0 there it is not considered a part of the number as it is a leading zero.

• 23 hours ago, # ^ | ← Rev. 3 →   0 Ig first they will go with number of problems with 100% acceptance rate, then quality of code. Then rest others.

• 23 hours ago, # ^ | I think they take various factors into account like your resume quality, college, CGPA, branch etc.

• 23 hours ago, # ^ | its based on points scored

• 4 hours ago, # ^ | I solved 3 and submitted 4th at the end (don't know if it was accepted or not as the test ended while compiling) and I got the clearance of Online Round mail. The ranking should be the only selection criteria I think which will be based on the points scored, in case of a tie, time will be considered.

  • 24 minutes ago, # ^ | ← Rev. 2 →   0 From Interview-Bit or from Trilogy? I have received one for successful completion from Interview-Bit. Don't know if it's in any way related to the actual selection process.

    • 13 minutes ago, # ^ | From Trilogy, for completing the aptitude test as the next round.

• 23 hours ago, # ^ | Manacher's Algo? Just found something new to learn. Thought of applying KMP or ZF in some way, but knew it wouldn't work.

• 20 hours ago, # ^ | ← Rev. 2 →   0 which one you did by prefix sum?the problem in which we need to give x1^x2 where x1 and x2 are bitwise and of range l1,r1,l2 and r2? If yes how?I did it using seg trees. upd got it from a below comment.

  • 111 minutes ago, # ^ | Simple method: We have to find the values of x1 and x2 then xor them. So, first we create a prefix array pref[i][j] of size n*32 which represents the sum of jth bit of the indices in the range [0, i]. Now, when we want to calculate value of x1, we find the number of set bits of the jth bit and check if it is equal to r1 — l1 + 1 (size), because for bitwise and to be 1, we need all the bits to be equal to 1. Similarly, we can find x1 and x2, then xor them

• 8 hours ago, # ^ | ← Rev. 3 →   0 Third problem can be also done by some simple string hashing and binary search. Make prefix and suffix hash using some mod value (I used 1e9+9). Now just binary search on the length uptil which the string [i,i+mid] from prefix hash and [i-mid,i] from suffix hash is same. This problem can be solved using binary search in log(n) time.

  • 5 hours ago, # ^ | can you please share some similar question like this?

• 23 hours ago, # ^ | Does Q representing the indexes i?

  • 23 hours ago, # ^ |

• 5 hours ago, # ^ | Just an addition, C was basically a direct implementation of Manacher's algo, which gives linear runtime.

• 4 hours ago, # ^ | did u solution for the problem c got ac i got tle for a few tc.My code get 286/300.

  • 3 hours ago, # ^ | This was my code for 3rd problem and I got an AC for that. def solve(s, b): n = len(s) s = '('+s+')' p = [0]*(n+1) l = 1 r = 1 for i in range(1, n+1): p[i] = max(0, min(r - i, p[l + (r - i)])) while(s[i-p[i]] == s[i+p[i]]): p[i] += 1 if(i + p[i] > r): l = i - p[i] r = i + p[i] ans = [] for i in b: ans.append(p[i]*2-1) return ans s = "abacaba" b = [2, 3, 4] print(solve(s, b))

    • 92 minutes ago, # ^ | thanks.

  • 95 minutes ago, # ^ | Yeah, my binary search + hashing solution gave 296/300 initially, then I removed the unecessary %mod operations during addition, i.e (a%m + b%m > m) ? (a%m + b%m — m) : (a%m + b%m). that was enough for my submission to then get a full score.

    • 92 minutes ago, # ^ | oh i see.Thanks for the reply..

• 75 minutes ago, # ^ | Can You Please share Your approach for problem C of binary search with hashing

• 16 minutes ago, # ^ | ← Rev. 2 →   0 In C, in test cases where the density of palindromes is high, wouldn't the effective complexity of Polynomial Hashing with Verification on Match be O(N*(LogN)*N)? The probability of hash collision is extremely low, I know, especially if a large prime is used, but still, without verification wouldn't it be non-deterministic?