Given a positive integer N, the task is to find the minimum number of operations required to reduce N to 1 by repeatedly dividing N by its proper divisors or by decreasing N by 1.
Examples:
Input: N = 9
Output: 3
Explanation:
The proper divisors of N(= 9) are {1, 3}. Following operations are performed to reduced N to 1:
Operation 1: Divide N(= 9) by 3(which is a proper divisor of N(= 9) modifies the value of N to 9/3 = 1.
Operation 2: Decrementing the value of N(= 3) by 1 modifies the value of N to 3 – 1 = 2.
Operation 3: Decrementing the value of N(= 2) by 1 modifies the value of N to 2 – 1 = 1.
Therefore, the total number of operations required is 3.Input: N = 4
Output: 2
Approach: The given problem can be solved based on the following observations:
- If the value of N is even, then it can be reduced to value 2 by dividing N by N / 2 followed by decrementing 2 to 1. Therefore, the minimum number of steps required is 2.
- Otherwise, the value of N can be made even by decrementing it and can be reduced to 1 using the above steps.
Follow the steps given below to solve the problem
- Initialize a variable, say cnt as 0, to store the minimum number of steps required to reduce N to 1.
- Iterate a loop until N reduces to 1 and perform the following steps:
- If the value of N is equal to 2 or N is odd, then update the value of N = N – 1 and increment cnt by 1.
- Otherwise, update the value of N = N / (N / 2) and increment cnt by 1.
- After completing the above steps, print the value of cnt 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 1int reduceToOne(long long int N){ // Stores the number // of steps required int cnt = 0; while (N != 1) { // If the value of N // is equal to 2 or N is odd if (N == 2 or (N % 2 == 1)) { // Decrement N by 1 N = N - 1; // Increment cnt by 1 cnt++; } // If N is even else if (N % 2 == 0) { // Update N N = N / (N / 2); // Increment cnt by 1 cnt++; } } // Return the number // of steps obtained return cnt;}// Driver Codeint main(){ long long int N = 35; cout << reduceToOne(N); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{ // Function to find the minimum number// of steps required to reduce N to 1static int reduceToOne(long N){ // Stores the number // of steps required int cnt = 0; while (N != 1) { // If the value of N // is equal to 2 or N is odd if (N == 2 || (N % 2 == 1)) { // Decrement N by 1 N = N - 1; // Increment cnt by 1 cnt++; } // If N is even else if (N % 2 == 0) { // Update N N = N / (N / 2); // Increment cnt by 1 cnt++; } } // Return the number // of steps obtained return cnt;}// Driver Codepublic static void main(String[] args) { long N = 35; System.out.println(reduceToOne(N));}}// This code is contributed by Dharanendra L V. |
Python3
# python program for the above approach# Function to find the minimum number# of steps required to reduce N to 1def reduceToOne(N): # Stores the number # of steps required cnt = 0 while (N != 1): # If the value of N # is equal to 2 or N is odd if (N == 2 or (N % 2 == 1)): # Decrement N by 1 N = N - 1 # Increment cnt by 1 cnt += 1 # If N is even elif (N % 2 == 0): # Update N N = N / (N / 2) # Increment cnt by 1 cnt += 1 # Return the number # of steps obtained return cnt# Driver Codeif __name__ == '__main__': N = 35 print (reduceToOne(N))# This code is contributed by mohit kumar 29. |
C#
// C# program for the above approachusing System; class GFG{// Function to find the minimum number// of steps required to reduce N to 1static int reduceToOne(long N){ // Stores the number // of steps required int cnt = 0; while (N != 1) { // If the value of N // is equal to 2 or N is odd if (N == 2 || (N % 2 == 1)) { // Decrement N by 1 N = N - 1; // Increment cnt by 1 cnt++; } // If N is even else if (N % 2 == 0) { // Update N N = N / (N / 2); // Increment cnt by 1 cnt++; } } // Return the number // of steps obtained return cnt;} // Driver Codepublic static void Main(){ long N = 35; Console.WriteLine(reduceToOne(N));}}// This code s contributed by code_hunt. |
Javascript
<script>// Javascript program for the above approach// Function to find the minimum number// of steps required to reduce N to 1function reduceToOne( N){ // Stores the number // of steps required let cnt = 0; while (N != 1) { // If the value of N // is equal to 2 or N is odd if (N == 2 || (N % 2 == 1)) { // Decrement N by 1 N = N - 1; // Increment cnt by 1 cnt++; } // If N is even else if (N % 2 == 0) { // Update N N = Math.floor(N / Math.floor(N / 2)); // Increment cnt by 1 cnt++; } } // Return the number // of steps obtained return cnt;}// Driver Codelet N = 35;document.write(reduceToOne(N));// This code is contributed by jana_sayantan.</script> |
Output:
3
Time Complexity: O(1)
Auxiliary Space: O(1)
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