# Digit Removal CodeChef Solution

Digit Removal CodeChef Solution: Hello coders, today we are going to solve Digit Removal CodeChef Solution which is part of CodeChef Solutions

As you already know that this site does not contain only the Codefchef solutions here, you can also find the solution for other programming problems. I.e. LeetcodeC programsC++ Programs SolutionsPython ProgramsWeb Technology, Data StructuresRDBMS Programs and Java Programs Solutions.

## Problem

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

## Input

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

## Output

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

## Constraints

• 1 ≤ T ≤ 105
• 1 ≤ N ≤ 109
• 0 ≤ D ≤ 9

Subtask 1 (100 points): Original constraints

## Example

```Input:
5
21 5
8 8
100 0
5925 9
434356 3
Output:
1
11
75
5644
```

## Explanation

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

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

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

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

### Digit Removal CodeChef Solution in JAVA

```/* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0)
{
int n=sc.nextInt();
int d=sc.nextInt();
int newn = n;
int f_digit,ans=0,p=0;
while( newn >0 ){
f_digit = newn % 10;
newn /=  10;
p++;
if(f_digit==d){
newn = newn * (int)Math.pow(10, p);
newn += (f_digit + 1) * (int)Math.pow(10, p-1);
ans = newn - n;
p =0;
}
}
System.out.println(ans);
}
}
}
```

### Digit Removal CodeChef Solution in CPP

```#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;
}```

### Digit Removal CodeChef Solution in Python

```# cook your dish here
def counting(idx, k, num, s, count):
for j in range(idx + 1, k):
count = count * 10 + (num - int(s[j]))
return count
def nine(a):
i=a.find('9')
if i==0:
return str(pow(10,len(a)))
else:
return a[:i-1]+str(int(a[i-1])+1)+('0'*len(a[i:len(a)]))
for _ in range(int(input())):
n, d = map(int, input().split())
s = str(n)
count = 0
if (n == d):
print("1")
else:
idx = s.find(str(d))
if (idx == -1):
print("0")
else:
k = len(s)
if (d == 0):
count = counting(idx, k, 10, s, 0)
print(count + 1)
else:
if (d!=9):
count = counting(idx, k, 9, s, 0)
print(count + 1)
else:
t=False
if '9' in s:
t=True
dup=s[:]
while(t==True):
dup=nine(dup)
if '9' not in dup:
t=False
print(int(dup)-n)
# cook your dish here
```

Disclaimer: The above Problem (Digit Removal CodeChef) is generated by CodeChef but the solution is provided by Chase2learn.This tutorial is only for Educational and Learning purpose.

Finally, we are now, in the end, I just want to conclude some important message for you

Note:- I compile all programs, if there is any case program is not working and showing an error please let me know in the comment section. If you are using adblocker, please disable adblocker because some functions of the site may not work correctly.

Please share our posts on social media platforms and also suggest to your friends to Join Our Groups. Don’t forget to subscribe.