Minimize given flips required to reduce N to 0

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Given an integer N, the task is to reduce the value of N to 0 by performing the following operations minimum number of times:

Flip the rightmost (0^th) bit in the binary representation of N.
If (i – 1)^th bit is set, then flip the i^th bit and clear all the bits from (i – 2)^th to 0^th bit.

Examples:

Input: N = 3
Output: 2
Explanation:
The binary representation of N (= 3) is 11
Since 0^th bit in binary representation of N(= 3) is set, flipping the 1^st bit of binary representation of N modifies N to 1(01).
Flipping the rightmost bit of binary representation of N(=1) modifies N to 0(00).
Therefore, the required output is 2

Input: N = 4
Output: 7

Approach: The problem can be solved based on the following observations:

1 -> 0 => 1
10 -> 11 -> 01 -> 00 => 2 + 1 = 3
100 -> 101 -> 111 -> 110 -> 010 -> … => 4 + 2 + 1 = 7
1000 -> 1001 -> 1011 -> 1010 -> 1110 -> 1111 -> 1101 -> 1100 -> 0100 -> … => 8 + 7 = 15
Therefore, for N = 2^N total (2^{(N + 1)} – 1) operations required.
If N is not a power of 2, then the recurrence relation is:
MinOp(N) = MinOp((1 << cntBit) – 1) – MinOp(N – (1 << (cntBit – 1)))
cntBit = total count of bits in binary representation of N.
MinOp(N) denotes minimum count of operations required to reduce N to 0.

Follow the steps below to solve the problem:

Calculate the count of bits in binary representation of N using log₂(N) + 1.
Use the above recurrence relation and calculate the minimum count of operations required to reduce N to 0.

Below is the implementation of the above approach.

C++

// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum count of
// operations required to Reduce N to 0
int MinOp(int N)
{
 
    if (N <= 1)
        return N;
 
    // Stores count of
    // bits in N
    int bit = log2(N) + 1;
 
    // Recurrence relation
    return ((1 << bit) - 1)
           - MinOp(N - (1 << (bit - 1)));
}
 
// Driver Code
int main()
{
 
    int N = 4;
    cout << MinOp(N);
    return 0;
}

Java

// Java program to implement 
// the above approach 
class GFG{
     
// Function to find the minimum count of 
// operations required to Reduce N to 0 
public static int MinOp(int N) 
{ 
    if (N <= 1) 
        return N; 
   
    // Stores count of 
    // bits in N 
    int bit = (int)(Math.log(N) / 
                    Math.log(2)) + 1; 
   
    // Recurrence relation 
    return ((1 << bit) - 1) - MinOp(
        N - (1 << (bit - 1))); 
} 
 
// Driver code
public static void main(String[] args)
{
    int N = 4; 
     
    System.out.println(MinOp(N));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python program to implement
# the above approach
 
# Function to find the minimum count of
# operations required to Reduce N to 0
import math
def MinOp(N):
    if (N <= 1):
        return N;
 
    # Stores count of
    # bits in N
    bit = (int)(math.log(N) / math.log(2)) + 1;
 
    # Recurrence relation
    return ((1 << bit) - 1) - MinOp(N - (1 << (bit - 1)));
 
# Driver code
if __name__ == '__main__':
    N = 4;
 
    print(MinOp(N));
 
# This code is contributed by 29AjayKumar 

C#

// C# program to implement
// the above approach  
using System;
 
class GFG{
     
// Function to find the minimum count of 
// operations required to Reduce N to 0 
public static int MinOp(int N) 
{ 
    if (N <= 1) 
        return N; 
         
    // Stores count of 
    // bits in N 
    int bit = (int)(Math.Log(N) / 
                    Math.Log(2)) + 1; 
                     
    // Recurrence relation 
    return ((1 << bit) - 1) - MinOp(
        N - (1 << (bit - 1))); 
} 
  
// Driver code
public static void Main()
{
    int N = 4; 
     
    Console.WriteLine(MinOp(N));
}
}
 
// This code is contributed by sanjoy_62

Javascript

<script>
// javascript program to implement
// the above approach
 
// Function to find the minimum count of
// operations required to Reduce N to 0
function MinOp(N)
{
    if (N <= 1)
        return N;
    
    // Stores count of
    // bits in N
    let bit = (Math.log(N) /
                    Math.log(2)) + 1;
    
    // Recurrence relation
    return ((1 << bit) - 1) - MinOp(
        N - (1 << (bit - 1)));
}
 
// Driver code
    let N = 4;
    document.write(MinOp(N));
     
    // This code is contributed by souravghosh0416.
</script>

Output:

Time Complexity: O(log(N)) //since the logarithm function is used, hence the time complexity is logarithmic
Auxiliary Space: O(1) // since no extra variable is used hence the space is taken by the algorithm is constant

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