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Minimum Factorization in C++
Suppose we have a positive integer x, we have to find the smallest positive integer b whose multiplication of each digit equals to x. If we have no such answer then return 0.
So, if the input is like 48, then the output will be 68
To solve this, we will follow these steps −
ret := 0, mul := 1
if a < 2, then:
return a
for initialize i := 9, when i >= 2, update (decrease i by 1), do −
while a mod i is same as 0, do −
ret := i * mul + ret
mul := mul * 10
a := a / i
return (if a < 2 and ret < inf, then ret, otherwise 0)
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
public:
int smallestFactorization(int a) {
lli ret = 0;
lli mul = 1;
if (a < 2)
return a;
for (lli i = 9; i >= 2; i--) {
while (a % i == 0) {
ret = i * mul + ret;
mul *= 10;
a /= i;
}
}
return a < 2 && ret < INT_MAX ? ret : 0;
}
};
main(){
Solution ob;
cout << (ob.smallestFactorization(48));
}Input
48
Output
68
Updated on: 16-Nov-2020
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