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C++ program to count minimum number of operations needed to make number n to 1
Suppose we have a number n. We perform any one of these operations arbitrary number of times −
Replace n with n/2 when n is divisible by 2
Replace n with 2n/3 when n is divisible by 3
Replace n with 4n/5 when n is divisible by 5
We have to count minimum number of moves needed to make number 1. If not possible, return -1.
So, if the input is like n = 10, then the output will be 4, because use n/2 to get 5, then 4n/5 to get 4, then n/2 again to get 2 and n/2 again to get 1.
Steps
To solve this, we will follow these steps −
m := 0 while n is not equal to 1, do: if n mod 2 is same as 0, then: n := n / 2 (increase m by 1) otherwise when n mod 3 is same as 0, then: n := n / 3 m := m + 2 otherwise when n mod 5 is same as 0, then: n := n / 5 m := m + 3 Otherwise m := -1 Come out from the loop return m
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
int solve(int n) {
int m = 0;
while (n != 1) {
if (n % 2 == 0) {
n = n / 2;
m++;
}
else if (n % 3 == 0) {
n = n / 3;
m += 2;
}
else if (n % 5 == 0) {
n = n / 5;
m += 3;
}
else {
m = -1;
break;
}
}
return m;
}
int main() {
int n = 10;
cout << solve(n) << endl;
}Input
10
Output
4
Updated on: 03-Mar-2022
249 Views
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