Digit Removal Solution - Codechef

 

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You are given an integer  N $N$  and a digit  D $D$ . Find the minimum non-negetive integer you should add to  N $N$  such that the final value of  N $N$  does not contain the digit  D $D$ .

Input Format

• The first line contains  T $T$  denoting the number of test cases. Then the test cases follow.
• Each test case contains two integers  N $N$  and  D $D$  on a single line denoting the original number and the digit you need to avoid.

Output Format

For each test case, output on a single line the minimum non-negetive integer you should add to  N $N$ .

Constraints

• 1≤T≤105 $1\le T\le {10}^5$
• 1≤N≤109 $1\le N\le {10}^9$
• 0≤D≤9 $0\le D\le 9$

Subtasks

• Subtask 1 (100 points): Original constraints

Sample Input 1 

 
5

21 5
8 8
100 0
5925 9
434356 3

Sample Output 1 

0
1
11
75
5644

Explanation

Test case  1 $1$ :  N=21 $N=21$  does not contain the digit  D=5 $D=5$ . Hence there is no need to add any integers to  N $N$ .

Test case  2 $2$ : If  1 $1$  is added to  N=8 $N=8$ , it becomes equal to  9 $9$ , which does not contain the digit  D=8 $D=8$ .

Test case  3 $3$ : The minimum integer you should add to  N=100 $N=100$  such that the final value of  N $N$  does not contain the digit  D=0 $D=0$  is  11 $11$ .

Test case  5 $5$ : The minimum integer which is greater than  434356 $434356$  and does not contain the digit  D=3 $D=3$  is  440000 $440000$ . So we should add  440000−434356=5644 $440000-434356=5644$ .

Digit Removal Solution: 

  
#include<bits/stdc++.h>
using namespace std;

int isMin(int n, int d){
    int newN = n, rem , count =0, c = 0;
    while(newN>0){
        rem = newN % 10;
        newN = newN /10;
        c++;
        if(rem == d){
            newN = newN*pow(10,c)+(rem+1)*pow(10,c-1);
            count = newN -n;
            c = 0;
        }
    }
    return count;
}

int main() 
{
  int t;
  cin>>t;
  while(t--){
      int n, d;
      cin>>n>>d;
      cout<<isMin(n, d) <<endl;
  }
  return 0;
}