Reduce a number to 1 by performing given operations
Given a number N. The task is to reduce the given number N to 1 in the minimum number of steps. You can perform any one of the below operations in each step.
- Operation 1: If the number is even then you can divide the number by 2.
- Operation 2: If the number is odd then you are allowed to perform either (n+1) or (n-1).
You need to print the minimum number of steps required to reduce the number N to 1 by performing the above operations.
Examples:
Input : n = 15
Output : 5
15 is odd 15+1=16
16 is even 16/2=8
8 is even 8/2=4
4 is even 4/2=2
2 is even 2/2=1
Input : n = 7
Output : 4
7->6
6->3
3->2
2->1
The idea is to recursively compute the minimum number of steps required.
- If the number is even, then we are allowed to only divide the number by 2.
- But, when the number is Odd, we can either increment or decrement it by 1. So, we will use recursion for both n-1 and n+1 and return the one with the minimum number of operations.
Below is the implementation of above approach:
C++
// CPP program to count minimum // steps to reduce a number #include <iostream> #include <cmath> using namespace std; int countways(int n) { if (n == 1) return 0; else if (n % 2 == 0) return 1 + countways(n / 2); else return 1 + min(countways(n - 1), countways(n + 1)); } // Driver code int main() { int n = 15; cout << countways(n)<< "\n"; return 0; } |
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Java
// Java program to count minimum // steps to reduce a number class Geeks { static int countways(int n) { if (n == 1) return 0; else if (n % 2 == 0) return 1 + countways(n / 2); else return 1 + Math.min(countways(n - 1), countways(n + 1)); } // Driver code public static void main(String args[]) { int n = 15; System.out.println(countways(n)); } } // This code is contributed by ankita_saini |
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Python3
# Python program to count minimum # steps to reduce a number def countways(n): if (n == 1): return 0; elif (n % 2 == 0): return 1 + countways(n / 2); else: return 1 + min(countways(n - 1), countways(n + 1)); # Driver code n = 15; print(countways(n)); # This code is contributed by PrinciRaj1992 |
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C#
// C# program to count minimum // steps to reduce a number using System; class GFG { static int countways(int n) { if (n == 1) return 0; else if (n % 2 == 0) return 1 + countways(n / 2); else return 1 + Math.Min(countways(n - 1), countways(n + 1)); } // Driver code static public void Main () { int n = 15; Console.Write(countways(n)); } } // This code is contributed by Raj |
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Output:
5
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