RPD - Editorial

PROBLEM LINK:

Div1
Div2
Practice

Setter- Hasan Jaddouh
Tester- Roman Bilyi
Editorialist- Abhishek Pandey

DIFFICULTY:

CAKEWALK

PRE-REQUISITES:

Implementation

PROBLEM:

Given an array A of size N, find the largest sum of digits obtainable from product of any two numbers of the array.

QUICK-EXPLANATION:

Key to AC- A look at the constraints and implementation skills!

Since the size of the array is only upto 100, there are \binom{N}{2}, i.e. \frac{N*(N-1)}{2} ways of choosing the elements. Hence we simply choose all possible pairs, find their product, find the sum of digits of the product, and report the maximum one out of those!

EXPLANATION:

The solution is pretty straightforward, hence we will discuss it in a single section, along with code for reference.

The first thing after taking input is to formulate what we are doing. What we will do is-

  1. Pick a pair out of the \binom{N}{2} pairs.
  2. Find its product and sum of digits.
  3. If this is greater than sum of digits seen so far, update answer appropriately.

First, we will have to pick the pair of elements we wish to multiply. This is achieved by using 2 nested loops-

for(i=0;i<n;i++)
    {
        for(j=i+1;j<n;j++)
        {
            int product = arr[i] * arr[j];
        }
    }

The next step, is to find the sum of digits. It is a good practice to make functions for any common functionality we reuse - and to keep code clean. Hence, we will make a function to find and return sum of digit of the given number. We will then call it from the loops we just created.

int sumOfDigits(int n)
{
    int sum=0;
    while(n>0)
    {
        sum+=n%10; // n%10 gives the digit at units place
        n/=10; // Integer division leads to loss of digit at unit's place, and shifts all digits by 1 place right.
    }
    return sum;
}

All that is left is to call this function, and update the answer appropriately. Our final code looks somewhat like this-

int ans=0;
    
    for(i=0;i<n;i++)
    {
        for(j=i+1;j<n;j++)
        {
            int product = arr[i] * arr[j];
            int sum= sumOfDigits(product);
            ans=max(ans,sum);
        }
    }

SOLUTION

Setter
 #include <iostream>
#include <algorithm>
#include <string>
#include <assert.h>
using namespace std;
 
 
 
long long readInt(long long l,long long r,char endd){
	long long x=0;
	int cnt=0;
	int fi=-1;
	bool is_neg=false;
	while(true){
		char g=getchar();
		if(g=='-'){
			assert(fi==-1);
			is_neg=true;
			continue;
		}
		if('0'<=g && g<='9'){
			x*=10;
			x+=g-'0';
			if(cnt==0){
				fi=g-'0';
			}
			cnt++;
			assert(fi!=0 || cnt==1);
			assert(fi!=0 || is_neg==false);
			
			assert(!(cnt>19 || ( cnt==19 && fi>1) ));
		} else if(g==endd){
			assert(cnt>0);
			if(is_neg){
				x= -x;
			}
			assert(l<=x && x<=r);
			return x;
		} else {
			assert(false);
		}
	}
}
string readString(int l,int r,char endd){
	string ret="";
	int cnt=0;
	while(true){
		char g=getchar();
		assert(g!=-1);
		if(g==endd){
			break;
		}
		cnt++;
		ret+=g;
	}
	assert(l<=cnt && cnt<=r);
	return ret;
}
long long readIntSp(long long l,long long r){
	return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
	return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
	return readString(l,r,'\n');
}
string readStringSp(int l,int r){
	return readString(l,r,' ');
}
 
int T;
int n;
int arr[111];
 
int sumdig(int x){
	if(x==0)return 0;
	return (x%10) + sumdig(x/10);
}
 
 
int main(){
	//freopen("00.txt","r",stdin);
	//freopen("00o.txt","w",stdout);
	T=readIntLn(1,100);
	while(T--){
		int sl = 0;
		n=readIntLn(1,100);
		for(int i=0;i<n;i++){
			if(i==n-1){
				arr[i]=readIntLn(1,10000);
			} else {
				arr[i]=readIntSp(1,10000);
			}
		}
		for(int i=0;i<n;i++){
			for(int j=i+1;j<n;j++){
				sl = max(sl, sumdig(arr[i] * arr[j]));
			}
		}
		cout<<sl<<endl;
	}
	assert(getchar()==-1);
} 
Tester
#include "bits/stdc++.h"
using namespace std;
 
#define FOR(i,a,b) for (int i = (a); i < (b); i++)
#define RFOR(i,b,a) for (int i = (b) - 1; i >= (a); i--)
#define ITER(it,a) for (__typeof(a.begin()) it = a.begin(); it != a.end(); it++)
#define FILL(a,value) memset(a, value, sizeof(a))
 
#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define PB push_back
#define MP make_pair
 
typedef long long Int;
typedef vector<int> VI;
typedef pair<int, int> PII;
 
const double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000;
const Int LINF = INF * (Int) INF;
const int MAX = 307; 
 
const int MOD = 998244353;
 
long long readInt(long long l,long long r,char endd){
    long long x=0;
    int cnt=0;
    int fi=-1;
    bool is_neg=false;
    while(true){
        char g=getchar();
        if(g=='-'){
            assert(fi==-1);
            is_neg=true;
            continue;
        }
        if('0'<=g && g<='9'){
            x*=10;
            x+=g-'0';
            if(cnt==0){
                fi=g-'0';
            }
            cnt++;
            assert(fi!=0 || cnt==1);
            assert(fi!=0 || is_neg==false);
           
            assert(!(cnt>19 || ( cnt==19 && fi>1) ));
        } else if(g==endd){
            assert(cnt>0);
            if(is_neg){
                x= -x;
            }
            assert(l<=x && x<=r);
            return x;
        } else {
            assert(false);
        }
    }
}
string readString(int l,int r,char endd){
    string ret="";
    int cnt=0;
    while(true){
        char g=getchar();
        assert(g!=-1);
        if(g==endd){
            break;
        }
        cnt++;
        ret+=g;
    }
    assert(l<=cnt && cnt<=r);
    return ret;
}
long long readIntSp(long long l,long long r){
    return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
    return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
    return readString(l,r,'\n');
}
string readStringSp(int l,int r){
    return readString(l,r,' ');
}
void assertEof(){
    assert(getchar()==-1);
}
 
int f(int n)
{
    int res = 0;
    while (n)
    {
        res += n % 10;
        n /= 10;
    }
    return res;
}
 
int main(int argc, char* argv[])
{
	// freopen("in.txt", "r", stdin);
	//ios::sync_with_stdio(false); cin.tie(0);
    
    int t = readIntLn(1, 100);
    FOR(tt,0,t)
    {
        int n = readIntLn(2, 100);
        VI A(n);
        FOR(i,0,n)
            A[i] = readInt(1, 10000, (i + 1 < n ? ' ' : '\n'));
        int res = 0;
        FOR(i,0,n)
            FOR(j,i + 1,n)
                res = max(res, f(A[i] * A[j]));
        cout << res << endl;
    }
 
    assertEof();
 
    cerr << 1.0 * clock() / CLOCKS_PER_SEC << endl;
    
} 
Editorialist
  /*
 *
 ********************************************************************************************
 * AUTHOR : Vijju123                                                                        *
 * Language: C++14                                                                          *
 * Purpose: -                                                                               *
 * IDE used: Codechef IDE.                                                                  *
 ********************************************************************************************
 *
 Comments will be included in practice problems if it helps ^^
 */
 
 
 
#include <iostream>
#include<bits/stdc++.h>
using namespace std;
 
//I never understand why people put useless functions at top of code. Makes it so unreadable ughhhh.
int mod=pow(10,9)+7;
int fastExpo(long long a,long long n, int mod)
{
    a%=mod;
    if(n==2)return a*a%mod;
    if(n==1)return a;
    if(n&1)return a*fastExpo(fastExpo(a,n>>1,mod),2,mod)%mod;
    else return fastExpo(fastExpo(a,n>>1,mod),2,mod);
}
inline void add(vector<vector<int> > &a,vector<vector<int> > &b,int mod)
{
    for(int i=0;i<a.size();i++)for(int j=0;j<a[0].size();j++)b[i][j]=(b[i][j]+a[i][j])%mod;
}
 
void multiply(vector<vector<int> > &a, vector<vector<int> > &b,int mod,vector<vector<int> > &temp)
{
    assert(a[0].size()==b.size());
    int i,j;
    for(i=0;i<a.size();i++)
    {
        for(j=0;j<b[0].size();j++)
        {
            temp[i][j]=0;
            for(int p=0;p<a[0].size();p++)
            {
                temp[i][j]=(temp[i][j]+1LL*a[i][p]*b[p][j])%mod;
            }
        }
    }
}
 
void MatExpo(vector<vector<int> > &arr,int power,int mod)
{
    int i,j,k;
    vector<vector<int> >temp,temp2,temp3;
    vector<int> init(arr[0].size());
    for(i=0;i<arr.size();i++){temp.push_back(init);}
    temp3=temp;
    temp2=temp;
    for(i=0;i<arr.size();i++)temp3[i][i]=1;
    while(power>0)
    {
        if(power&1)
        {
            multiply(arr,temp3,mod,temp);
            swap(temp3,temp);
        }
        multiply(arr,arr,mod,temp2);
        swap(arr,temp2);
        power>>=1;
    }
    swap(arr,temp3);
}
 
long long readInt(long long l,long long r,char endd){
	long long x=0;
	int cnt=0;
	int fi=-1;
	bool is_neg=false;
	while(true){
		char g=getchar();
		if(g=='-'){
			assert(fi==-1);
			is_neg=true;
			continue;
		}
		if('0'<=g && g<='9'){
			x*=10;
			x+=g-'0';
			if(cnt==0){
				fi=g-'0';
			}
			cnt++;
			assert(fi!=0 || cnt==1);
			assert(fi!=0 || is_neg==false);
 
			assert(!(cnt>19 || ( cnt==19 && fi>1) ));
		} else if(g==endd){
			if(is_neg){
				x= -x;
			}
			assert(l<=x && x<=r);
			return x;
		} else {
			assert(false);
		}
	}
}
string readString(int l,int r,char endd){
	string ret="";
	int cnt=0;
	while(true){
		char g=getchar();
		assert(g!=-1);
		if(g==endd){
			break;
		}
		cnt++;
		ret+=g;
	}
	assert(l<=cnt && cnt<=r);
	return ret;
}
long long readIntSp(long long l,long long r){
	return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
	return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
	return readString(l,r,'\n');
}
string readStringSp(int l,int r){
	return readString(l,r,' ');
}
 
vector<int> primes;
int isComposite[1000001]={1,1};
void sieve()
{
    int i,j;
    for(i=2;i<=1000000;i++)
    {
        if(!isComposite[i])
        {
            primes.push_back(i);
            isComposite[i]=i;
        }
        for(j=0;j<primes.size() and i*primes[j]<=1000000;j++)
        {
            isComposite[primes[j]*i]=i;
            if(i%primes[j]==0)break;
        }
    }
}
 
int sumOfDigits(int n)
{
    int sum=0;
    while(n>0)
    {
        sum+=n%10;
        n/=10;
    }
    return sum;
}
 
int main() {
	// your code goes here
	#ifdef JUDGE
	string in,out;
	cin>>in>>out;
	in+=".in";
	if(in!="z.in")
        out+=".out";
    else out+=".in";
	if(in!="z.in")
    freopen(in.c_str(), "rt", stdin);
    freopen(out.c_str(), "wt", stdout);
    #endif
	ios_base::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
	srand(time(NULL));
	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	
	int t;
	cin>>t;
	while(t--)
	{
	    int n;
	    cin>>n;
	    int arr[n],i,j,k;
	    for(i=0;i<n;i++)cin>>arr[i];
	    int ans=0;
	    
	    for(i=0;i<n;i++)
	    {
	        for(j=i+1;j<n;j++)
	        {
	            int product = arr[i] * arr[j];
	            int sum= sumOfDigits(product);
	            ans=max(ans,sum);
	        }
	    }
	    cout<<ans<<endl;
	}
	
	assert(getchar()==-1);
	return 0;
}

Time Complexity=O(N^2LogA_i)
Space Complexity=O(N)

CHEF VIJJU’S CORNER :smiley:

1.Discuss solution if constraints of A_i are raised upto 10^{18}. Does the solution change at all?
2. Discuss solution if constraints of A_i are raised upto 10^{300}. How will you solve it without using Big Integer libraries?

Hint

Take input of A_i as a string!! Or, explore ways by which you can declare an integer array, which takes only 1 digit in input.

3.How many steps does it take to find sum of digit of a number \leq 10^4 ? Answer the same for finding sum of digit of a number \leq 10^{300}

Answer

5 and 301 respectively. In general, if there are N digits in a number, it takes O(N) time to find sum of digits!

1 Like

plz, tell me what is wrong with my code.

test=int(input())
for i in range(0,test):
	n=int(input())
	t=[]
	m=[]
	t=list(map(int,input().split()))
	for i in range(n):
	    if i+1==n:
	        x=t[i]*t[0]
	        m.append(x)
	    else:
	        x=t[i]*t[i+1]
	        m.append(x)
	    
	x=list(sum(map(int, str(i))) for i in m)
	print(max(x))

isn’t the time complexity is O(N^2logAi) ?

What’s wrong in my code?

def SumofDigits(n):
    su=0
    while(n!=0):
        su+=n%10
        n=n/10
    return su

if __name__=='__main__':
    for x in range(int(input())):
        m=int(input())
        a=list(map(int,input().split()))
        ans=0
        for i in range(m):
            if i<m-1:
                for j in range(i+1,m):
                    product=int(a[i]*a[j])
                    su=int(SumofDigits(product))
                    ans=max(ans,su)
        print(ans)

Yup, my bad - missed fixing the typo! :frowning:

These are my favourite kind of problems​:grin::smile::sweat_smile:

1 Like

Is there any approach with o(n) complexity?

in your SumOfDigits function change n=n/10 to n=n//10, ‘//’ is used in python for integer division.