Minimum steps to reach the Nth stair in jumps of perfect power of 2

Given N stairs, the task is to find the minimum number of jumps of perfect power of 2 requires to reach the Nth stair.

Examples:

Input: N = 5
Output:
Explanation:
We can take jumps from 0->1->4.
So the minimum jumps require are 2.

Input: N = 23
Output: 4
Explanation:
We can take jumps from 0->1->2->4->16.
So the minimum jumps required are 4.

Approach: Since the jumps are required to be in perfect power of 2. So the count of set bit in the given number N is the minimum number of jumps required to reach Nth stair as the summation of 2i for all set bit index i is equals to N.



Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include "bits/stdc++.h"
using namespace std;
  
// Function to count the number of jumps
// required to reach Nth stairs.
int stepRequired(int N)
{
  
    int cnt = 0;
  
    // Till N becomes 0
    while (N) {
  
        // Removes the set bits from
        // the right to left
        N = N & (N - 1);
        cnt++;
    }
  
    return cnt;
}
  
// Driver Code
int main()
{
  
    // Number of stairs
    int N = 23;
  
    // Function Call
    cout << stepRequired(N);
    return 0;
}

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Java

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// Java program for the above approach
  
import java.util.*;
  
class GFG{
  
// Function to count the number of jumps
// required to reach Nth stairs.
static int stepRequired(int N)
{
  
    int cnt = 0;
  
    // Till N becomes 0
    while (N > 0) {
  
        // Removes the set bits from
        // the right to left
        N = N & (N - 1);
        cnt++;
    }
  
    return cnt;
}
  
// Driver Code
public static void main(String[] args)
{
  
    // Number of stairs
    int N = 23;
  
    // Function Call
    System.out.print(stepRequired(N));
}
}
  
// This code is contributed by PrinciRaj1992

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Python3

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# Python3 program for the above approach
  
# Function to count the number of jumps
# required to reach Nth stairs.
def stepRequired(N):
  
    cnt = 0;
  
    # Till N becomes 0
    while (N > 0):
  
        # Removes the set bits from
        # the right to left
        N = N & (N - 1);
        cnt += 1;
    return cnt;
  
# Driver Code
if __name__ == '__main__':
  
    # Number of stairs
    N = 23;
  
    # Function Call
    print(stepRequired(N));
      
# This code is contributed by 29AjayKumar

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C#

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// C# program for the above approach
using System;
  
class GFG{
  
// Function to count the number of 
// jumps required to reach Nth stairs.
static int stepRequired(int N)
{
    int cnt = 0;
  
    // Till N becomes 0
    while (N > 0)
    {
  
        // Removes the set bits from
        // the right to left
        N = N & (N - 1);
        cnt++;
    }
  
    return cnt;
}
  
// Driver Code
public static void Main(String[] args)
{
  
    // Number of stairs
    int N = 23;
  
    // Function Call
    Console.Write(stepRequired(N));
}
}
  
// This code is contributed by 29AjayKumar

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Output:

4

Time Complexity: O(log N)

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Improved By : princiraj1992, 29AjayKumar