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

23 месяца назад, # ^ |

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.

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23 месяца назад, # ^ |

I also hate the IDE of InterviewBit. Worst experience I have with the IDE.

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ShivanshJ

23 месяца назад, # ^ |

+18

Exactly its soo hard to debug there!

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23 месяца назад, # |

← Rev. 2 →

First Problem___

You are given a range l,r. We have to count total no of beautiful numbers in that range.

Numbers are beautiful if, xor of all digits of that number is greater than the avg(max digit of that number + min digit of that number)

return result%1e9+7

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23 месяца назад, # ^ |

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.

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23 месяца назад, # ^ |

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.

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23 месяца назад, # ^ |

Did u get ac for this?

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23 месяца назад, # ^ |

yes

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angry_pikachu

23 месяца назад, # ^ |

can you share the solution of 6-D dp?

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KaI777

23 месяца назад, # ^ |

do you have any idea what will be the final cutoff for the shortlisting process?

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23 месяца назад, # ^ |

Could you please share the DP approach?

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onlyerror

23 месяца назад, # ^ |

← Rev. 3 →

Mine is 6D Dp. Could not find the bug in contest time Later found it.

class Solver{
    final long MOD=(long) 1e9+7;
    String s;
    long [][][][][][]dp;
    Solver(String s){
        this.s=s;
        this.dp=new long[2][2][10][10][16][20];
        for(long [][][][][]x:dp){
            for(long [][][][]xx:x){
                for(long [][][]xxx:xx){
                    for(long [][] xxxx:xxx){
                        for(long []xxxxx:xxxx){
                            Arrays.fill(xxxxx,-1);
                        }
                    }
                }
            }
        }
    }
    public long solve(int index,int smalltaken,int nonzero,int mind,int maxd,int xorv){
        if(index==s.length()){
            return (nonzero==1&&2*xorv>(mind+maxd))?1:0;
        }
        if(dp[nonzero][smalltaken][mind][maxd][xorv][index]!=-1){
           return dp[nonzero][smalltaken][mind][maxd][xorv][index];
        }
        int range=(smalltaken==1)?9:s.charAt(index)-'0';
        long ans=0;
        for(int d=0;d<=range;d++){
            int n_nonzero=(nonzero==1||d>0)?1:0;
            int n_mind=(n_nonzero==1)?Math.min(d,mind):mind;
            int n_smalltaken=(smalltaken==1||d<range)?1:0;
            ans+=solve(index+1,n_smalltaken,n_nonzero,n_mind,Math.max(d,maxd),xorv^d);
            ans%=MOD;
        }
        return dp[nonzero][smalltaken][mind][maxd][xorv][index]=ans;
    }
}
class Test{
    public static void main(String[] args) {
        // A -> 170888 B -> 676277
        Solver solve1=new Solver("170887");//use A - 1 where A is the smaller number
        Solver solve2=new Solver("676277");
        System.out.println(solve2.solve(0,0,0,9,0,0)-solve1.solve(0,0,0,9,0,0));
    }
}

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angry_pikachu

23 месяца назад, # ^ |

← Rev. 2 →

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;
}

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23 месяца назад, # ^ |

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.

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Snapper_001

23 месяца назад, # ^ |

What was the constraint over l and r

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23 месяца назад, # ^ |

(1, 10^18)

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23 месяца назад, # |

Btw, what is the selection criteria for the Online Round?
Is it based on the number of problems solved, or quality of submissions, or something else?
There wasn't even a rankings page, afaik.

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23 месяца назад, # ^ |

← Rev. 3 →

Ig first they will go with number of problems with 100% acceptance rate, then quality of code. Then rest others.

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23 месяца назад, # ^ |

I think they take various factors into account like your resume quality, college, CGPA, branch etc.

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lamellama

23 месяца назад, # ^ |

its based on points scored

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hackerbhaiya

23 месяца назад, # ^ |

+10

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.

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23 месяца назад, # ^ |

← Rev. 2 →

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.

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hackerbhaiya

23 месяца назад, # ^ |

From Trilogy, for completing the aptitude test as the next round.

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aKILLL

23 месяца назад, # |

first was digit dp, second was usual prefix sum, third I did using Manachers algo.

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23 месяца назад, # ^ |

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

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23 месяца назад, # ^ |

← Rev. 2 →

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.

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k_rishabh

23 месяца назад, # ^ |

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

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___LoneWolf___

23 месяца назад, # ^ |

← Rev. 3 →

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.

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onlyerror

23 месяца назад, # ^ |

can you please share some similar question like this?

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23 месяца назад, # |

3rd Problem

Given a string s and a vector Q containing queries. For each element i in Q we have to find the length of longest odd palindrome whose middle index is i in string s. 1 <= Len(s), Len(Q) <= 1e5. We have to return vector of results of each query

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mr_golem

23 месяца назад, # ^ |

Does Q representing the indexes i?

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https://leetcode.com/discuss/interview-question/2924881/intern-hunt-challenge-trilogy-innovations-2023/

23 месяца назад, # ^ |

yes

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__iOTA

23 месяца назад, # |

In second problem just store prefix of every bit for all element and check if bit[r][j] — bit[l-1][j] == r-l+1...if yes add that bit to answer. Calculate the same for both given range and store the xor of both!!

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Tony_Kakkar

23 месяца назад, # |

+13

Here are the problems:

problem 1

problem 2

problem 3

problem 4

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mr_golem

23 месяца назад, # ^ |

how did u get them

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Neothermal

23 месяца назад, # ^ |

telegram xd:

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Tony_Kakkar

23 месяца назад, # ^ |

+23

Don't directly jump to conclusions idiot!

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Tony_Kakkar

23 месяца назад, # ^ |

From here:

https://leetcode.com/discuss/interview-question/2924881/intern-hunt-challenge-trilogy-innovations-2023/

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coderdhanraj

23 месяца назад, # |

+15

Here is my approaches...

Problem A

Problem B

Problem C

Problem D

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vrintle

23 месяца назад, # ^ |

Just an addition, C was basically a direct implementation of Manacher's algo, which gives linear runtime.

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23 месяца назад, # ^ |

did u solution for the problem c got ac i got tle for a few tc.My code get 286/300.

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sangeeta

23 месяца назад, # ^ |

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))

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23 месяца назад, # ^ |

thanks.

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yeetholmes619

23 месяца назад, # ^ |

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.

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23 месяца назад, # ^ |

oh i see.Thanks for the reply..

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tandonjiii

23 месяца назад, # ^ |

Can You Please share Your approach for problem C of binary search with hashing

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coderdhanraj

23 месяца назад, # ^ |

Sure! so I just hashed the string in both the directions i.e. forward and backward/reverse. Then for a query of 'i' I searched for the maximum length that has the center 'i' and is odd. i.e. maximum x for [i — x... i + x] is palindrome. you may check palindrome nature by taking hash between [i — x .... i] for reverse hash table and [i ... i + x] for forward hash table and just compare the result. They should be equal for a substring being palindrome!

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tandonjiii

23 месяца назад, # ^ |

thanks

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