CMX1P03 - Code To Get Her - Code Mates x1 Editorial



Author & Editorialist : Mayuresh Patle
Tester : Uttkarsh Bawankar




Combinations, Counting, Modular Arithmetic, Modular Multiplicative Inverse.


You are given a string S containing lowercase English alphabets. You are also given a list of N names (which are strings too, containing only lowercase English alphabets). For each of these names, you’ve to calculate the total number of sub-sequences of string S which are anagrams of that name. Since this value could be very large, calculate this value modulo 998244535. This value is called as the score of the name.

Finally, print the name having maximum score, if there are more than one such names, then print the one which appeared first in the list. If all names have 0 score, then print “No Research This Month” (without quotes).

You’ve to do this for M testcases.


The score of each name is, \prod ^{S.count(c)}C_{name.count(c)} for each unique character c in name. Here, ^nC_r represents the total number of ways to select r objects out of available n objects.


By definition, if a two strings are anagrams of each other, then the frequency of occurrences of each unique character in both of the string must be same. For example, if two string A and B are anagrams of each other, and the character c occurs exactly x times in A then, c must occurs exactly x times in B and vice-versa.

Let’s denote the count of occurrences of character c in string A by A.count(c). Now, we want to count the number of sub-sequences of S which satisfy the the above condition. Hence,

  • We can first count and store the frequency of occurrences of each of the 26 alphabets in S.
  • Then, for each name in the list, we will store the count of occurrences of each alphabet.
  • Now, for each character c, we need to select exactly name.count(c) occurrences out of available S.count(c) occurrences in S. Total number of ways to select these for character c is ^{S.count(c)}C_{name.count(c)}.
  • Hence, the final score will be the product of above expression for each c \in lowercase alphabets.

Some points to mention:

  • The calculations might lead to overflow, hence, apply modulo operation after each step.
  • After applying modulo operation, direct division will give wrong results, hence to calculate \frac{A}{B} \mod 998244353, you will have to calculate (A\ * B^{-1}) \mod 9982443535. Here, B^{-1} is the modular multiplicative inverse of B under modulo 998244353.

Time Complexity : O(|S| + N*(|name|+26*log_2(9982443535) )) for each testcase.


C++ Solution
using namespace std;
typedef long long ll;
typedef vector <ll> vll;

const ll md=998244353, maxN=100001;

//function for modular exponentiation, returns (n^p) mod md
ll modex(ll n,ll p)
    ll ans=1;
        if(p&1) ans=(ans*n)%md;
    return ans;

ll fact[maxN]={1};

/*This function returns nCr mod md, i.e { n!/((n-r)!*r!) } mod md
nCr = no. of ways to select r objects out of n available objects
This function works in O(log2(md)) time*/
ll ncr(ll n,ll r)
    if(n<r) return 0;
    ll num=fact[n], den=(fact[n-r]*fact[r])%md;
    ll inv=modex(den,md-2);
    return (num*inv)%md;

int main()
    //fast io

    ll M,N,i,maxscore;
    string S,name,result;

    for(i=1;i<maxN;++i) fact[i]=(i*fact[i-1])%md;  //calculate factorials

        result="No Research This Month";

        vll cntS(26);
        for(char c:S) ++cntS[c-'a'];

            ll score=1;
            vll cntname(26);

            for(char c:name) ++cntname[c-'a'];
            for(i=0;score and i<26;++i)                     //calculate score
            cout<<score<<"\n";                              //print score

            if(score>maxscore)                              //update result
        cout<<result<<"\n";                                 //print result
    return 0;
Python 3 Solution
def ncr(n,r):                  #takes O(r*log2(md)) time
    for i in range(r):
    return (num*inv)%md
for _ in range(int(input())):
    for c in a: cnts[c]=0
    for c in s: cnts[c]+=1
    maxscore,result=0,"No Research This Month"
    for _ in range(int(input())):
        for c in a: cnt[c]=0
        for c in name: cnt[c]+=1
        for c in a:
            if score==0: break

        if score>maxscore:

I read the discussion title and thought it was related to human trafficking :rofl: :rofl:


Hi, I didn’t have much time today so couldn’t really spend time on the contest unfortunately : /

Could you tell me why this gives TLE?

import numpy as np 
SIZE = 26
def binomialCoeff(n, k): 
	res = 1
	if (k > n - k): k = n - k  
	for i in range(k): res *= (n - i);res = int(res/(i + 1)) 
	return res
def countSubsequences(str1, str2):
	freq1 = np.zeros(26, dtype =;freq2 = np.zeros(26, dtype = 
	n1 = len(str1);n2 = len(str2)
	for i in range(n1): freq1[ord(str1[i]) - ord('a') ] += 1
	for i in range(n2): freq2[ord(str2[i]) - ord('a')] += 1
	count = 1
	for i in range(SIZE):
		if (freq2[i] != 0):
			if (freq2[i] <= freq1[i]): count = count * binomialCoeff(freq1[i],freq2[i])
			else: return 0
	return count
for _ in range(int(input())):
	S = input();ans = {}
	for _ in range(int(input())):
		name = input()
		ans[name] = countSubsequences(S,name)
	if(max(ans.values()) > 0): print(max(ans,key = ans.get))
	else: print("No Research This Month") #sample testcases passed, TLE
def past(arr,W,i):arr[i - 1] += W;return arr
def present(arr,L,R): print(sum(arr[L - 1:R]))
N = int(input()); arr = list(map(int , input().split()))
for _ in range(int(input())):
	input_ = list(map(str , input().split()))
	if(input_[0] == "past"): past(arr,int(input_[1]),int(input_[2]))
	else: present(arr,int(input_[1]),int(input_[2])) #sample testcases passed, TLE

even this code got TLE for the ‘dark’ problem

In this function, the calculations are being done on large numbers, which is taking considerable amount of time. Hence, resulting in TLE.
You’ll have to use modular arithmetic. Please read the editorial.

1 Like

okay thank you

For Dark, the expected time complexity was O(Q*LogN). Editorial for Dark is available here.

1 Like

lol :joy:

1 Like

Can anybody help me why my code is wrong?
Link :
I have tried to follow the same logic as editorial but can’t seem to find my mistake

editorial for ?

@sb_kmb Editorial for Code Together (CMX1P01): CMX1P01 - Code Together - Code Mates x1 Editorial

1 Like

@mayureshpatle can you please point out the mistake in my code :

I have used the same C++ code given at geeksforgeeks with some changes :

may be i am doing mistake with applying the mod…please help…any help is greatly appreciated

@nalingoyal In function binomialCoefficient, the values of n and k are large, so there will be an overflow during calculation. Use the method mentioned in the editorial.

1 Like

@mayureshpatle i did as u suggested…i checked for overflow in binomial cofficient function…but i didnt get AC …plz have a look.

@nalingoyal \frac{A}{B} \mod M \neq \frac{A \mod M}{B \mod M}.

\frac{A}{B} \mod M = ((A \mod M)*B^{-1}) \mod M, here B^{-1} is the modular multiplicative inverse of B under modulo M.

To use this, you will have to calculate \text{numerator} \mod M and \text{denominator} \mod M differently. Also, since M=998244353 is prime, you can use Fermat’s Little Theorem to calculate the modular multiplicative inverse of B under modulo M, which will be equal to B^{M-2} \mod M.

Hence, the result will be:

((\text{numerator} \mod M) * ((\text{denominator} \mod M)^{M-2}\mod M)) \mod M

Have a look at the ncr() function in Python 3 Solution provided in the editorial, it uses the same approach that you are trying to implement.

Also, you can read more about Modular multiplicative inverse here.

1 Like

it is not resolved yet.plz help
lli power(lli x, unsigned long long int y, unsigned long long int mod)
if (y == `0)
return 1;
long long int p = power(x, y/2, mod) % mod;
p = (p * p) % mod;

return (y%2 == 0)? p : (x * p) % mod; 

// Returns value of Binomial
// Coefficient C(n, k)
lli binomialCoeff(lli n, lli k)
lli res = 1,den=1;

// Since C(n, k) = C(n, n-k) 
if (k > n - k) 
	k = n - k; 

// Calculate value of 
// [n * (n-1) *---* (n-k+1)] / 
// [k * (k-1) *----* 1] 
for (lli i = 0; i < k; ++i) { 
	res *= (n - i);
            res %= mod;
	       den %=mod;
lli inv=power(den,mod-2,mod)%mod;
return (res*inv)%mod;


my solution:

@redindian I had to spend a lot of time on debugging your code. And, finally found the bug. The problem is with this statement:

result = (result*(fact[x[i]]*binaryExp((fact[c[i]]*fact[x[i]-c[i]])%mod,mod-2))%mod)%mod;

Let’s make it look a little simpler,

Let, a = fact[x[i]]
b = (fact[c[i]]*fact[x[i]-c[i]])%mod
c = binaryExp(b, mod-1)

Now, put these symbols in your statement

result = (result * (a * c) % mod)% mod;

Now you can guess the error. :relieved:

result * (a * c) % mod has same execution sequence as (result * a * c) % mod, and each of result, a & c can have values upto 998244353, hence, this will result in long long limit overflow.

The correct statement should be of the following form:

result = (result * ( (a * c) % mod ) )% mod;

Just, replace the statement, with the following one, and your code will be accepted.

result = (result*((fact[x[i]]*binaryExp((fact[c[i]]*fact[x[i]-c[i]])%mod,mod-2))%mod))%mod;

You see, the problem was just because of a pair of parentheses, and it was quite tough to spot the error, so I would suggest, rather than writing a large expression in a single statement, you should always evaluate it by dividing into simpler forms. This will make your code more readable and easy to debug. :slightly_smiling_face:

1 Like

You’re violating this condition

print the name having maximum score, if there are more than one such names, then print the one which appeared first in the list.

P.S. There’s one more error in your code,

lli inv=power(den,mod-2,mod)%mod;
return (res*inv)%mod;

These two statements should be placed after the closing bracket, i.e outside the loop’s body, you’ve written them inside the loop’s body.

1 Like

thanks man… i got my mistake…actually i thought Unorderd_map keeps the thing in the order they are inserted…but that was not correct…thanks for helping me…
people like you are the reason why i like codechef discussion forum so much :smiling_face_with_three_hearts:

1 Like

Thank you very much :smile: