Data Modelling & AI Data Structure & Algorithm

Minimize steps to change N to power of 2 by deleting or appending any digit

24 June 2025

0

Given an integer N, the task is to find the minimum number of steps required to change the number N to a perfect power of 2 using the following steps:

Delete any one digit d from the number.
Add any digit d at the end of the number N.

Examples:

Input: N = 1092
Output: 2
Explanation:
Following are the steps followed:

Removing digit 9 from the number N(= 1092) modifies it to 102.

Adding digit 4 to the end of number N(= 102) modifies it to 1024.

After the above operations, the number N is converted into perfect power of 2 and the number of moves required is 2, which is the minimum. Therefore, print 2.

Input: N = 4444
Output: 3

Approach: The given problem can be solved using the Greedy Approach, the idea is to store all the numbers that are the power of 2 and are less than 10²⁰ and check with each number the minimum steps to convert the given number into a power of 2. Follow the steps below to solve the problem:

Initialize an array and store all the numbers that are the power of 2 and less than 10²⁰.
Initialize the answer variable best as length + 1 as the maximum number of steps needed.
Iterate over the range [0, len) where len is the length of the array using the variable x and perform the following steps:
- Initialize the variable, say position as 0.
- Iterate over the range [0, len) where len is the length of the number using the variable i and if the position is less than len(x) and x[position] is equal to num[i], then increase the value of a position by 1.
- Update the value of best as the minimum of best or len(x) + len(num) – 2*position.
After performing the above steps, print the value of best as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include<bits/stdc++.h> 
using namespace std;
 
// Function to find the minimum number of
// steps required to reduce N to perfect
// power of 2 by performing given steps
void findMinimumSteps(int N){
 
    string num = to_string(N);
    long long c = 1;
    stack<string>a;
 
    // Stores all the perfect power of 2
    while(true){
        if (c > pow(10,20) || c>10)
            break;
        a.push(to_string(c));
        c = c * 2;
    }
 
    // Maximum number of steps required
    long long best = num.length() + 1;
     
    // Iterate for each perfect power of 2
    while(a.empty() == false){
        string x = a.top();
        a.pop();
        long long position = 0;
         
        // Comparing with all numbers
        for(int i=0;i<num.length();i++){
         
            if(position < x.length() && x[position] == num[i])
                position += 1;
 
        }
                 
        // Update the minimum number of
        // steps required
        best = min(best, (int)x.length() + (int)num.length() - 2 * position);
 
    }
         
    // Print the result
    cout<<(best-1)<<endl;
}
 
// Driver Code
int main()
{
    int N = 1092;
 
    // Function Call
    findMinimumSteps(N);
}
 
// code is contributed by shinjanpatra

Java

// Java Program to implement
// the above approach 
import java.util.*;
 
class GFG {
 
  // Function to find the minimum number of
  // steps required to reduce N to perfect
  // power of 2 by performing given steps
  static void findMinimumSteps(int N) {
    String num = String.valueOf(N);
    int c = 1;
 
    Stack<String> a = new Stack<>();
 
    // Stores all the perfect power of 2
    while (true) {
      if (c > (int)Math.pow(10, 20)|| c>10 ) {
        break;
      }
      a.add(String.valueOf(c));
      c = c * 2;
    }
 
    // Maximum number of steps required
    int best = num.length()+1;
 
    // Iterate for each perfect power of 2
    String x;
    int i = 0;
    while (!a.isEmpty()) {
      x = a.pop();
      int position = 0;
 
      // Comparing with all numbers
      for (i = 0; i < num.length(); i++) {
 
        if (position < x.length() && x.charAt(position) == num.charAt(i))
          position += 1;
      }
 
      // Update the minimum number of
      // steps required
      best = (int) (Math.min(best, x.length() + num.length() - 2 * position));
 
    }
 
    // Print the result
    System.out.print(best-1);
 
  }
 
  // Driver Code
  public static void main(String[] args) {
    int N = 1092;
 
    // Function Call
    findMinimumSteps(N);
  }
}
 
// This code is contributed by Rajput-Ji 

Python3

# Python program for the above approach
 
# Function to find the minimum number of
# steps required to reduce N to perfect
# power of 2 by performing given steps
def findMinimumSteps(N):
 
    num = str(N)
    c = 1
    a = []
 
    # Stores all the perfect power of 2
    while True:
        if (c > 10 ** 20):
            break
        a.append(str(c))
        c = c * 2
 
    # Maximum number of steps required
    best = len(num) + 1
     
    # Iterate for each perfect power of 2
    for x in a:
        position = 0
         
        # Comparing with all numbers
        for i in range(len(num)):
           
            if position < len(x) and x[position] == num[i]:
                position += 1
                 
        # Update the minimum number of
        # steps required
        best = min(best, len(x) + len(num) - 2 * position)
         
    # Print the result
    print(best)
 
# Driver Code
N = 1092
 
# Function Call
findMinimumSteps(N)

C#

// C# Program to implement
// the above approach
using System;
using System.Collections;
 
class GFG {
 
// Function to find the minimum number of
// steps required to reduce N to perfect
// power of 2 by performing given steps
static void findMinimumSteps(int N) {
    string num = N.ToString();
    int c = 1;
 
    Stack a = new Stack();
 
    // Stores all the perfect power of 2
    while (true) {
    if (c > Math.Pow(10, 20)|| c>10 ) {
        break;
    }
    a.Push(c.ToString());
    c = c * 2;
    }
 
    // Maximum number of steps required
    int best = num.Length+1;
 
    // Iterate for each perfect power of 2
    string x;
    int i = 0;
    while (a.Count!=0) {
    x = a.Peek().ToString();
    a.Pop();
    int position = 0;
 
    // Comparing with all numbers
    for (i = 0; i < num.Length; i++) {
 
        if (position < x.Length && x[position] == num[i])
        position += 1;
    }
 
    // Update the minimum number of
    // steps required
    best = Math.Min(best, x.Length + num.Length - 2 * position);
 
    }
 
    // Print the result
    Console.Write(best-1);
 
}
 
// Driver Code
public static void Main() {
    int N = 1092;
 
    // Function Call
    findMinimumSteps(N);
}
}
 
// This code is contributed by Pushpesh Raj

Javascript

<script>
 
        // JavaScript Program to implement
        // the above approach 
 
        // Function to find the minimum number of
        // steps required to reduce N to perfect
        // power of 2 by performing given steps
        function findMinimumSteps(N)
        {
            num = N.toString()
            c = 1
            a = []
 
            // Stores all the perfect power of 2
            while (1) {
                if (c > Math.pow(10, 20)) {
                    break;
                }
                a.push(c.toString())
                c = c * 2
            }
             
            // Maximum number of steps required
            best = num.length + 1
 
            // Iterate for each perfect power of 2
            for (x of a) {
                position = 0
 
                // Comparing with all numbers
                for (let i = 0; i < num.length; i++) {
 
                    if (position < x.length && x[position] == num[i])
                        position += 1
                }
                 
                // Update the minimum number of
                // steps required
                best = Math.min(best, x.length + num.length - 2 * position)
            }
             
            // Print the result
            document.write(best)
        }
         
        // Driver Code
        let N = 1092
 
        // Function Call
        findMinimumSteps(N)
 
    // This code is contributed by Potta Lokesh
    </script>

Output:

Time Complexity: O(N)
Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

Minimize steps to change N to power of 2 by deleting or appending any digit

C++

Java

Python3

C#

Javascript

7 Years, 2 Major Rebuilds, 40K+ GitHub Stars: The Rise of Milvus as the Leading Open-Source Vector Database

Gemini 3 Pro + Milvus: Building a More Robust RAG With Advanced Reasoning and Multimodal Power

Multimodal RAG Made Simple: RAG-Anything + Milvus Instead of 20 Separate Tools

LEAVE A REPLY Cancel reply

Most Popular

What’s going on with the Galaxy S26 Ultra? [Video]

Mobile’s next big thing is unraveling, it’s going to get messy, and it may already be too late to stop it

I used the Moto G 2026 and Samsung Galaxy A16, and only one makes sense for $200

Samsung’s Galaxy S26 series may offer a couple of software surprises

EDITOR PICKS

What’s going on with the Galaxy S26 Ultra? [Video]

Mobile’s next big thing is unraveling, it’s going to get messy, and it may already be too late to stop it

I used the Moto G 2026 and Samsung Galaxy A16, and only one makes sense for $200

POPULAR POSTS

What’s going on with the Galaxy S26 Ultra? [Video]

Mobile’s next big thing is unraveling, it’s going to get messy, and it may already be too late to stop it

I used the Moto G 2026 and Samsung Galaxy A16, and only one makes sense for $200

POPULAR CATEGORY

ABOUT US

FOLLOW US