Given an integer N. The task is to reduce the given number N to 1 in minimum number of given operations. You can perform any one of the below operations in each step.
- If the number is even then you can divide the number by 2.
- If the number is odd then you are allowed to perform either (N + 1) or (N – 1).
The task is 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 = 4
Output: 2
Approach: A recursive approach to solve the above problem has already been discussed in this article. In this article, an even optimised approach will be discussed.
The first step towards the solution is to realize that you’re allowed to remove the LSB only if it’s zero i.e. the operation of the first type. Now, what about the odd numbers. One may think that you just need to remove as many 1’s as possible to increase the evenness of the number which is not correct, for example:
111011 -> 111010 -> 11101 -> 11100 -> 1110 -> 111 -> 1000 -> 100 -> 10 -> 1
And yet, this is not the best way because
111011 -> 111100 -> 11110 -> 1111 -> 10000 -> 1000 -> 100 -> 10 -> 1
Both 111011 -> 111010 and 111011 -> 111100 remove the same number of 1’s, but the second way is better.
So, maximum number of 1’s have to be removed, doing +1 in case of a tie will fail for the testcase when n = 3 because 11 -> 10 -> 1 is better than 11 -> 100 -> 10 -> 1. Fortunately, that’s the only exception.
So the logic is:
- If N is even.
- Perform the first operation i.e. division by 2.
- If N is odd.
- If N = 3 or (N – 1) has less number of 1’s than (N + 1).
- Decrement N.
- else
- Increment N.
- If N = 3 or (N – 1) has less number of 1’s than (N + 1).
Below is the implementation of the above approach:
CPP
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the number// of set bits in nint set_bits(int n){ int count = 0; while (n) { count += n % 2; n /= 2; } return count;}// Function to return the minimum// steps required to reach 1int minSteps(int n){ int ans = 0; while (n != 1) { // If n is even then divide it by 2 if (n % 2 == 0) n /= 2; // If n is 3 or the number of set bits // in (n - 1) is less than the number // of set bits in (n + 1) else if (n == 3 or set_bits(n - 1) < set_bits(n + 1)) n--; else n++; // Increment the number of steps ans++; } // Return the minimum number of steps return ans;}// Driver codeint main(){ int n = 15; cout << minSteps(n); return 0;} |
Java
// Java implementation of the approachclass GFG{// Function to return the number// of set bits in nstatic int set_bits(int n){ int count = 0; while (n > 0) { count += n % 2; n /= 2; } return count;}// Function to return the minimum// steps required to reach 1static int minSteps(int n){ int ans = 0; while (n != 1) { // If n is even then divide it by 2 if (n % 2 == 0) n /= 2; // If n is 3 or the number of set bits // in (n - 1) is less than the number // of set bits in (n + 1) else if (n == 3 || set_bits(n - 1) < set_bits(n + 1)) n--; else n++; // Increment the number of steps ans++; } // Return the minimum number of steps return ans;}// Driver codepublic static void main(String[] args){ int n = 15; System.out.print(minSteps(n));}}// This code is contributed by PrinciRaj1992 |
Python
# Python3 implementation of the approach# Function to return the number# of set bits in ndef set_bits(n): count = 0 while (n): count += n % 2 n //= 2 return count# Function to return the minimum# steps required to reach 1def minSteps(n): ans = 0 while (n != 1): # If n is even then divide it by 2 if (n % 2 == 0): n //= 2 # If n is 3 or the number of set bits # in (n - 1) is less than the number # of set bits in (n + 1) elif (n == 3 or set_bits(n - 1) < set_bits(n + 1)): n -= 1 else: n += 1 # Increment the number of steps ans += 1 # Return the minimum number of steps return ans# Driver coden = 15print(minSteps(n))# This code is contributed by mohit kumar 29 |
C#
// C# implementation of the approachusing System;class GFG{// Function to return the number// of set bits in nstatic int set_bits(int n){ int count = 0; while (n > 0) { count += n % 2; n /= 2; } return count;}// Function to return the minimum// steps required to reach 1static int minSteps(int n){ int ans = 0; while (n != 1) { // If n is even then divide it by 2 if (n % 2 == 0) n /= 2; // If n is 3 or the number of set bits // in (n - 1) is less than the number // of set bits in (n + 1) else if (n == 3 || set_bits(n - 1) < set_bits(n + 1)) n--; else n++; // Increment the number of steps ans++; } // Return the minimum number of steps return ans;}// Driver codepublic static void Main(String[] args){ int n = 15; Console.Write(minSteps(n));}}// This code is contributed by Rajput-Ji |
Javascript
<script>// JavaScript implementation of the approach// Function to return the number// of set bits in nfunction set_bits(n){ let count = 0; while (n) { count += n % 2; n = parseInt(n / 2); } return count;}// Function to return the minimum// steps required to reach 1function minSteps(n){ let ans = 0; while (n != 1) { // If n is even then divide it by 2 if (n % 2 == 0) n = parseInt(n / 2); // If n is 3 or the number of set bits // in (n - 1) is less than the number // of set bits in (n + 1) else if (n == 3 || set_bits(n - 1) < set_bits(n + 1)) n--; else n++; // Increment the number of steps ans++; } // Return the minimum number of steps return ans;}// Driver code let n = 15; document.write(minSteps(n));</script> |
5
Time Complexity: O(n)
Auxiliary Space: O(1)
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