2 Keys Keyboard Problem

0 0 4 minutes read

Given a positive integer N and a string S initially it is “A”, the task is to minimize the number of operations required to form a string consisting of N numbers of A’s by performing one of the following operations in each step:

Copy all the characters present in the string S.
Append all the characters to the string S which are copied last time.

Examples:

Input: N = 3
Output: 3
Explanation:
Below are the operations performed:
Operation 1: Copy the initial string S once i.e., “A”.
Operation 2: Appending the copied string i.e., “A”, to the string S modifies the string S to “AA”.
Operation 3: Appending the copied string i.e., “A”, to the string S modifies the string S to “AAA”.
Therefore, the minimum number of operations required to get 3 A’s is 3.

Input: N = 7
Output: 7

Approach: The given problem can be solved based on the following observations:

If N = P₁*P₂*P_m where {P₁, P₂, …, P_m} are the prime numbers then one can perform the following moves:
1. First, copy the string and then paste it (P₁ – 1) times.
2. Similarly, again copying the string and pasting it for (P₂ – 1) times.
3. Performing M times with the remaining prime number, one will get the string with N number of A’s.
Therefore, the total number of minimum moves needed is given by (P₁ + P₂ + … + P_m).

Follow the steps below to solve the problem:

Initialize a variable, say ans as 0, that stores the resultant number of operations.
Find the prime factors and its power of the given integer N.
Now, iterate through all the prime factors of N and increment the value of ans by the product of the prime factor and its power.
Finally, after completing the above steps, print the value of ans as the resultant minimum number of moves.

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 form N number
// of A's
int minSteps(int N)
{
    // Stores the count of steps needed
    int ans = 0;
 
    // Traverse over the range [2, N]
    for (int d = 2; d * d <= N; d++) {
 
        // Iterate while N is divisible
        // by d
        while (N % d == 0) {
 
            // Increment the value of
            // ans by d
            ans += d;
 
            // Divide N by d
            N /= d;
        }
    }
 
    // If N is not 1
    if (N != 1) {
        ans += N;
    }
 
    // Return the ans
    return ans;
}
 
// Driver Code
int main()
{
    int N = 3;
    cout << minSteps(N);
 
    return 0;
}

Java

// Java Program for the above approach
import java.io.*;
 
class GFG
{
   
    // Function to find the minimum number
    // of steps required to form N number
    // of A's
    static int minSteps(int N)
    {
        // Stores the count of steps needed
        int ans = 0;
 
        // Traverse over the range [2, N]
        for (int d = 2; d * d <= N; d++) {
 
            // Iterate while N is divisible
            // by d
            while (N % d == 0) {
 
                // Increment the value of
                // ans by d
                ans += d;
 
                // Divide N by d
                N /= d;
            }
        }
 
        // If N is not 1
        if (N != 1) {
            ans += N;
        }
 
        // Return the ans
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 3;
        minSteps(N);
        System.out.println(minSteps(N));
    }
}
 
// This code is contributed by lokesh potta.

Python3

# Python3 program for the above approach
# Function to find the minimum number
# of steps required to form N number
# of A's
def minSteps( N):
 
    # Stores the count of steps needed
    ans = 0
 
    # Traverse over the range [2, N]
    d = 2
    while(d * d <= N): 
 
        # Iterate while N is divisible
        # by d
        while (N % d == 0):
 
            # Increment the value of
            # ans by d
            ans += d
 
            # Divide N by d
            N /= d
        d += 1
     
    # If N is not 1
    if (N != 1): 
        ans += N
     
    # Return the ans
    return ans
 
# Driver Code
N = 3
print(minSteps(N))
 
# This code is contributed by shivanisinghss2110    

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the minimum number
// of steps required to form N number
// of A's
static int minSteps(int N)
{
    // Stores the count of steps needed
    int ans = 0;
 
    // Traverse over the range [2, N]
    for (int d = 2; d * d <= N; d++) {
 
        // Iterate while N is divisible
        // by d
        while (N % d == 0) {
 
            // Increment the value of
            // ans by d
            ans += d;
 
            // Divide N by d
            N /= d;
        }
    }
 
    // If N is not 1
    if (N != 1) {
        ans += N;
    }
 
    // Return the ans
    return ans;
}
 
// Driver Code
static public void Main ()
{
     
    int N = 3;
    Console.Write(minSteps(N));
}
}
 
// This code is contributed by sanjoy_62.

Javascript

<script>
 
// JavaScript Program for the above approach
// Function to find the minimum number
    // of steps required to form N number
    // of A's
    function minSteps(N)
    {
        // Stores the count of steps needed
        var ans = 0;
 
        // Traverse over the range [2, N]
        for (var d = 2; d * d <= N; d++) {
 
            // Iterate while N is divisible
            // by d
            while (N % d == 0) {
 
                // Increment the value of
                // ans by d
                ans += d;
 
                // Divide N by d
                N /= d;
            }
        }
 
        // If N is not 1
        if (N != 1) {
            ans += N;
        }
 
        // Return the ans
        return ans;
    }
 
    // Driver Code
        var N = 3;
        minSteps(N);
        document.write(minSteps(N));
   
// This code is contributed by shivanisinghss2110
 
</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 zambiatek!