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 (0th) bit in the binary representation of N.
- If (i – 1)th bit is set, then flip the ith bit and clear all the bits from (i – 2)th to 0th bit.
Examples:
Input: N = 3
Output: 2
Explanation:
The binary representation of N (= 3) is 11
Since 0th bit in binary representation of N(= 3) is set, flipping the 1st 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 2Input: 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 = 2N 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 0int 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 Codeint 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 codepublic 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 0import mathdef 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 codeif __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 codepublic 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 0function 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> |
7
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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