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*Pm where {P1, P2, …, Pm} are the prime numbers then one can perform the following moves:
- First, copy the string and then paste it (P1 – 1) times.
- Similarly, again copying the string and pasting it for (P2 – 1) times.
- 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 + … + Pm).
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'sint 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 Codeint main(){ int N = 3; cout << minSteps(N); return 0;} |
Java
// Java Program for the above approachimport 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'sdef 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 CodeN = 3print(minSteps(N))# This code is contributed by shivanisinghss2110 |
C#
// C# program for the above approachusing System;class GFG{ // Function to find the minimum number// of steps required to form N number// of A'sstatic 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 Codestatic 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
3
Time Complexity: O(?N)
Auxiliary Space: O(1)
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