Count Steps to Reduce N to 1 in C++

We are given a number N. The goal is to count the number of steps required to reduce the number to 1 by following rules −

• If the number is power of 2, reduce it to its half.

• Else reduce it to the N-(nearest power of 2 which is less than N).

For step 1, we will check if N is power of 2, by checking if ceil(log2(N)), floor(log2(N)) return the same result. If yes then N=N/3, increment count of operation.

If the result of step 1 is false then we will perform step 2 and subtract the nearest power of 2 less than N from N. Nearest power of 2 less than N will be calculated as −

x=floor(log2(N)) → when N is not power of 2, log2(N) gives floating point value, floor reduces it to the nearest integer less than N.

Now N=N-pow(2,x) → pow(2,x) will give the nearest power of 2 less than N. Reduce N.

Let’s understand with examples.

Input − N=20

Output-: Count of steps required − 3

Explanation − N=20

20 is not power of 2. Step 2. Reduce nearest power of 2 less than N from N. N=20- 16=4. Count=1.
4 is power of 2. Step 1. Reduce N to its half. N=4/2=2. Count=2.
2 is power of 2. Step 1. Reduce N to its half. N=2/2=1. Count=3.
N is 1 total step count=3.

Input − N=32

Output Count of steps required − 5

Explanation − N=32

32 is power of 2. Step 1. Reduce N to its half. N=32/2=16. Count=1.
16 is power of 2. Step 1. Reduce N to its half. N=16/2=8. Count=2.
8 is power of 2. Step 1. Reduce N to its half. N=8/2=4. Count=3.
4 is power of 2. Step 1. Reduce N to its half. N=4/2=2. Count=4.
2 is power of 2. Step 1. Reduce N to its half. N=2/2=1. Count=5.
N is 1 total step count=5.

Approach used in the below program is as follows

• We take an integer N for storing an integer value.

• Function stepCount(int n) takes N and returns the count of steps required to reduce it to 1.

• Take the initial count of steps as 0.

• Now while(n!=1) perform both steps 1, and 2 according to the value of n.

• If n is power of 2 ( ceil(log2(n))==floor(log2(n)) will be true ), reduce n to half. Increment count.

• If not power of 2 then reduce n by pow(2,x) where x is floor(log2(n)). Increment count.

• When the loop will be over then count will have the number of operations performed.

• Return count as desired result.

Example

 Live Demo

#include <iostream>
#include <math.h>
using namespace std;
// Function to return number of steps for reduction
int stepCount(int n){
   int count=0;
   while(n!=1){
      if(ceil(log2(n))==floor(log2(n))) //if n is power of 2 then this is true{
         n=n/2; //reduce n to half
         count++;
      } else {
         int x=floor(log2(n)); //floor value
         n=n-(pow(2,x)); //2^x is nearest power of 2 which is less than n
         count++;
      }
   }
   return count;
}
int main(){
   int N = 96;
   cout <<"Count of steps required to reduce N to 1:"<<stepCount(N);
   return 0;
}

Output

If we run the above code it will generate the following output −

Count of steps required to reduce N to 1:6