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