## ceaser

BAN USER- 0 Answers
**minimize digitsum**You are given positive integers N and D. You may perform operations of the following two types:

- ceaser October 19, 2018

add D to N, i.e. change N to N+D

change N to digitsum(N)

Here, digitsum(x) is the sum of decimal digits of x. For example, digitsum(123)=1+2+3=6, digitsum(100)=1+0+0=1, digitsum(365)=3+6+5=14.

You may perform any number of operations (including zero) in any order. Please find the minimum obtainable value of N and the minimum number of operations required to obtain this value.

this is my code I am getting the wrong answer in some test cases please let me know where I am wrong.

#include<iostream>

#include<queue>

using namespace std;

void check(long long int a, long long int &b, long long int &c, long long int d)

{

if(a<b)

{

b = a;

c = d;

}

}

int digitSum(long long int a)

{

int num = 0;

while(a>0)

{

num += a%10;

a = a/10;

}

return num;

}

void func(int N, int D, long long int &minElement, long long int &minTrial)

{

queue<long long int> q;

q.push(N);

if(N==1)

{

minElement = 1;

minTrial = 0;

}

long long int countNodes = q.size();

long long int level = 1;

int cnt = 1;

while(!q.empty())

{

countNodes = q.size();

while(countNodes>0)

{

long long int temp = q.front();

q.pop();

long long int first = temp+D;

check(first,minElement,minTrial,level);

if(minElement == 1)

{

return;

}

q.push(first);

long long second = digitSum(temp);

check(second,minElement,minTrial,level);

if(minElement == 1)

{

return;

}

q.push(second);

countNodes--;

cnt++;

if(cnt>1000000)

{

break;

}

}

if(cnt>1000000)

{

break;

}

level++;

}

}

int main()

{

long long int t = 1;

cin>>t;

while(t--)

{

long long int N, D;

cin>>N>>D;

long long int minElement = 1e18, minTrial = 1e18;

func(N,D,minElement, minTrial);

cout<<minElement<<" "<<minTrial<<endl;

}

return 0;

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