Given an integer N, the task is to find the number of steps required to reduce the given number N to 1 by performing the following operations:
- If the number is a power of 2, then divide the number by 2.
- Otherwise, subtract the greatest power of 2 smaller than N from N.
Examples:
Input: N = 2
Output: 1
Explanation: The given number can be reduced to 1 by following the following steps:
Divide the number by 2 as N is a power of 2 which modifies the N to 1.
Therefore, the N can be reduced to 1 in only 1 step.Input: N = 7
Output: 2
Explanation: The given number can be reduced to 1 by following the following steps:
Subtract 4 the greatest power of 2 less than N. After the step the N modifies to N = 7-4 = 3.
Subtract 2 the greatest power of 2 less than N. After the step the N modifies to N = 3-2 = 1.
Therefore, the N can be reduced to 1 in 2 steps.
Method 1 –
Approach: The idea is to recursively compute the minimum number of steps required.
- If the number is a power of 2, then we are allowed to only divide the number by 2.
- Else we can subtract the greatest power of 2 smaller than N from N.
- So, we will use recursion for both the operations whenever valid and return number of operations.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Utility function to check // if n is power of 2int highestPowerof2(int n){ int p = (int)log2(n); return (int)pow(2, p);}// Utility function to find highest// power of 2 less than or equal // to given numberbool isPowerOfTwo(int n){ if(n==0) return false; return (ceil(log2(n)) == floor(log2(n)));}// Recursive function to find // steps needed to reduce// a given integer to 1int reduceToOne(int N){ // Base Condition if(N == 1){ return 0; } // If the number is a power of 2 if(isPowerOfTwo(N) == true){ return 1 + reduceToOne(N/2); } // Else subtract the greatest //power of 2 smaller than N //from N else{ return 1 + reduceToOne(N - highestPowerof2(N)); }}// Driver Codeint main(){ // Input int N = 7; // Function call cout << reduceToOne(N) << endl;} |
Java
// java program for the above approachclass GFG { // Utility function to check // if n is power of 2 static int highestPowerof2(int n) { int p = (int)(Math.log(n) / Math.log(2)); return (int)Math.pow(2, p); } // Utility function to find highest // power of 2 less than or equal // to given number static boolean isPowerOfTwo(int n) { if(n==0) return false; return (int)(Math.ceil((Math.log(n) / Math.log(2)))) == (int)(Math.floor(((Math.log(n) / Math.log(2))))); } // Recursive function to find // steps needed to reduce // a given integer to 1 static int reduceToOne(int N) { // Base Condition if(N == 1){ return 0; } // If the number is a power of 2 if(isPowerOfTwo(N) == true){ return 1 + reduceToOne(N/2); } // Else subtract the greatest //power of 2 smaller than N //from N else{ return 1 + reduceToOne(N - highestPowerof2(N)); } } // Driver Code public static void main(String [] args) { // Input int N = 7; // Function call System.out.println(reduceToOne(N)); } }// This code is contributed by ihritik |
Python
# Python program for the above approachimport math# Utility function to check# Log base 2def Log2(x): if x == 0: return false; return (math.log10(x) / math.log10(2)); # Utility function to check# if x is power of 2def isPowerOfTwo(n): return (math.ceil(Log2(n)) == math.floor(Log2(n)));# Utility function to find highest# power of 2 less than or equal # to given numberdef highestPowerof2(n): p = int(math.log(n, 2)); return int(pow(2, p));# Recursive function to find # steps needed to reduce# a given integer to 1def reduceToOne(N): # Base Condition if(N == 1): return 0 # If the number is a power of 2 if(isPowerOfTwo(N) == True): return 1 + reduceToOne(N/2) # Else subtract the greatest # power of 2 smaller than N # from N else: return 1 + reduceToOne(N - highestPowerof2(N))# Driver Code# InputN = 7;#Function callprint(reduceToOne(N))# This code is contributed by Samim Hossain Mondal. |
C#
// C# program for the above approachusing System;class GFG { // Utility function to check // if n is power of 2 static int highestPowerof2(int n) { int p = (int)(Math.Log(n) / Math.Log(2)); return (int)Math.Pow(2, p); } // Utility function to find highest // power of 2 less than or equal // to given number static bool isPowerOfTwo(int n) { if(n == 0) return false; return (int)(Math.Ceiling((Math.Log(n) / Math.Log(2)))) == (int)(Math.Floor(((Math.Log(n) / Math.Log(2))))); } // Recursive function to find // steps needed to reduce // a given integer to 1 static int reduceToOne(int N) { // Base Condition if(N == 1){ return 0; } // If the number is a power of 2 if(isPowerOfTwo(N) == true){ return 1 + reduceToOne(N/2); } // Else subtract the greatest //power of 2 smaller than N //from N else{ return 1 + reduceToOne(N - highestPowerof2(N)); } } // Driver Code public static void Main() { // Input int N = 7; // Function call Console.Write(reduceToOne(N)); }}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script>// Javascript program for the above approach// Utility function to check // if n is power of 2function isPowerOfTwo(n){ if (n == 0) return false; return parseInt( (Math.ceil((Math.log(n) / Math.log(2))))) == parseInt( (Math.floor(((Math.log(n) / Math.log(2))))));}// Utility function to find highest// power of 2 less than or equal // to given numberfunction highestPowerof2(n){ let p = parseInt(Math.log(n) / Math.log(2), 10); return Math.pow(2, p);}// Recursive function to find // steps needed to reduce// a given integer to 1function reduceToOne(N){ // Base Condition if(N == 1){ return 0; } // If the number is a power of 2 if(isPowerOfTwo(N) == true){ return 1 + reduceToOne(N/2); } // Else subtract the greatest //power of 2 smaller than N //from N else{ return 1 + reduceToOne(N - highestPowerof2(N)); }}// Driver Code// Inputlet N = 7;// Function calldocument.write(reduceToOne(N));// This code is contributed by Samim Hossain Mondal</script> |
Output
2
Method 2 –
Approach: The given problem can be solved using bitwise operators. Follow the steps below to solve the problem:
- Initialize two variables say res with value 0 and MSB with value log2(N) to store the number of steps needed to reduce the number to 1 and the most significant bit of N.
- Iterate while N is not equal to 1:
- Check if the number is the power of 2 then divide N by 2.
- Otherwise, subtract the largest power of 2 less than N from N i.e update N as N=N-2MSB.
- Increment res by 1 and decrement MSB by 1.
- Finally, print res.
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 steps needed to// reduce a given integer to 1int reduceToOne(int N){ // Stores the most // significant bit of N int MSB = log2(N); // Stores the number of steps // required to reduce N to 1 int res = 0; // Iterates while N // is not equal 1 while (N != 1) { // Increment res by 1 res++; // If N is power of 2 if (N & (N - 1) == 0) { // Divide N by 2 N /= 2; } // Otherwise else { // Subtract 2 ^ MSB // from N N -= (1 << MSB); } // Decrement MSB by 1 MSB--; } // Returns res return res;}// Driver codeint main(){ // Input int N = 7; // Function call cout << reduceToOne(N) << endl;} |
Java
/*package whatever //do not write package name here */import java.io.*;class GFG { public static int logarithm(int number, int base) { int res = (int)(Math.log(number) / Math.log(base)); return res; } public static int reduceToOne(int N) { // Stores the most // significant bit of N int MSB = logarithm(N, 2); // Stores the number of steps // required to reduce N to 1 int res = 0; // Iterates while N // is not equal 1 while (N != 1) { // Increment res by 1 res++; // If N is power of 2 if ((N & (N - 1)) == 0) { // Divide N by 2 N /= 2; } // Otherwise else { // Subtract 2 ^ MSB // from N N -= (1 << MSB); } // Decrement MSB by 1 MSB--; } // Returns res return res; } public static void main(String[] args) { int N = 7; int res = reduceToOne(N); System.out.println(res); }}// This code is contributed by lokeshpotta20. |
Python3
# Python program for the above approachimport math# Function to find steps needed to# reduce a given integer to 1def reduceToOne(N): # Stores the most # significant bit of N MSB = math.floor(math.log2(N)) # Stores the number of steps # required to reduce N to 1 res = 0 # Iterates while N # is not equal 1 while (N != 1): # Increment res by 1 res += 1 # If N is power of 2 if (N & (N - 1) == 0): # Divide N by 2 N //= 2 # Otherwise else: # Subtract 2 ^ MSB # from N N -= (1 << MSB) # Decrement MSB by 1 MSB-=1 # Returns res return res # Driver code# InputN = 7# Function callprint(reduceToOne(N))# This code is contributed by shubham Singh |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to find steps needed to// reduce a given integer to 1static int reduceToOne(int N){ // Stores the most // significant bit of N int MSB = (int)(Math.Log(N)/Math.Log(2)); // Stores the number of steps // required to reduce N to 1 int res = 0; // Iterates while N // is not equal 1 while (N != 1) { // Increment res by 1 res++; // If N is power of 2 if ((N & (N - 1)) == 0) { // Divide N by 2 N /= 2; } // Otherwise else { // Subtract 2 ^ MSB // from N N -= (1 << MSB); } // Decrement MSB by 1 MSB--; } // Returns res return res;}// Driver codepublic static void Main(){ // Input int N = 7; // Function call Console.Write(reduceToOne(N));}}// This code is contributed by bgangwar59. |
Javascript
<script>// JavaScript program for the above approach function logarithm(number, base) { let res = (Math.log(number) / Math.log(base)); return res; } function reduceToOne(N) { // Stores the most // significant bit of N let MSB = logarithm(N, 2); // Stores the number of steps // required to reduce N to 1 let res = 0; // Iterates while N // is not equal 1 while (N != 1) { // Increment res by 1 res++; // If N is power of 2 if ((N & (N - 1)) == 0) { // Divide N by 2 N /= 2; } // Otherwise else { // Subtract 2 ^ MSB // from N N -= (1 << MSB); } // Decrement MSB by 1 MSB--; } // Returns res return res; } // Driver Code let N = 7; let res = reduceToOne(N); document.write(res);</script> |
Output
2
Time Complexity: O(log(N))
Auxiliary Space: O(1)
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