Closest Divisors in C++

Suppose we have an integer num, we have to find the closest two integers in absolute difference whose product equals num + 1 or num + 2. We have to find the two integers in any order. So if the input is 8, then the output will be [3, 3], for num + 1, it will be 9, the closest divisors are 3 and 3, for num + 2 = 10, the closest divisors are 2 and 5, hence 3 and 3 are chosen.

To solve this, we will follow these steps −

• Define a method called getDiv(), this will take x as input

• diff := infinity, create an array called ret of size 2

• for i := 1, if i^2 <= x, then increase i by 1

• if x is divisible by i, then

• a := i

• b := x / i

• newDiff := |a – b|

• if newDiff < diff, then

• diff := newDiff

• ret[0] := a and ret[1] := b

• return ret

• From the main method find op1 := getDiv(num + 1) and op2 := getDiv(num + 2)

• return op1 when |op1[0] – op[1]| <= |op2[0] – op2[1]|, otherwise op2

Example (C++)

Let us see the following implementation to get better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
class Solution {
public:
vector <int> getDiv(int x){
int diff = INT_MAX;
vector <int> ret(2);
for(int i = 1; i * i <= x; i++){
if(x % i == 0){
int a = i;
int b = x / i;
int newDiff = abs(a - b);
if(newDiff < diff){
diff = newDiff;
ret[0] = a;
ret[1] = b;
}
}
}
return ret;
}
vector<int> closestDivisors(int num) {
vector <int> op1 = getDiv(num + 1);
vector <int> op2 = getDiv(num + 2);
return abs(op1[0] - op1[1]) <= abs(op2[0] - op2[1]) ? op1 : op2;
}
};
main(){
Solution ob;
print_vector(ob.closestDivisors(8));
}

Input

8

Output

[3,3]

Updated on: 29-Apr-2020

280 Views