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Remove 9 in C++
Suppose we have an integer n, we have to return nth integer after following this operation: Start from integer 1, remove any integer that contains 9 such as 9, 19, 29... So now, we will have a new integer sequence like 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, ... We have to keep in mind that 1 will be the first integer.
So, if the input is like 9, then the output will be 10
To solve this, we will follow these steps −
ret := 0
s := 1
while n is non-zero, do −
ret := ret + (n mod 9) * s
n := n / 9
s := s * 10
s := s * 10
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
public:
int newInteger(int n) {
int ret = 0;
lli s = 1;
while (n) {
ret += (n % 9) * s;
n /= 9;
s *= 10;
}
return ret;
}
};
main(){
Solution ob;
cout << (ob.newInteger(120));
}Input
120
Output
143
Updated on: 11-Jul-2020
162 Views
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