Set bits in N equals to M in the given range.

3.3

You are given two 32-bit numbers, N and M, and two bit positions, i and j. Write a method to set all bits between i and j in N equal to M (e.g., M becomes a substring of N located at i and starting at j).

Examples:

Input : N = 1, M = 2, i = 2, j = 4
Output: 9
N = 00000001(Considering 8 bits only)
M = 10 (Binary of 2) For more indexes,
leading zeroes will be considered.
Now set 3 bits from ith index to j in 
the N as in the M.
Bits:-    0 0 0 (0  1  0) 0 1 = 9
Indexes:- 7 6 5  4  3  2  1 0
From index 2 to 4, bits are set according 
to the M.


Asked in : Adobe

A simple solution is to traverse all bits in N from 0 to 31 and set the bits equals to M in the range from i to j.

An efficient solution is to do following steps.

  1. Set all the bits after j in a number.
  2. Set all the bits before i in a number.
  3. Then perform Bitwise Or on both then we get the number with all the bits set except from i to j.
  4. Perform Bitwise And with the given N as to set the bits according to the N.
  5. Then shift M into the correct position i.e. in the range of i to j.
  6. And at the last perform Bitwise Or on (Shifted M and the N modifed in 4th step).
  7. The result will be N with M as substring from ith to jth bits

C++

// C++ program for above implementation
#include <iostream>
using namespace std;

// Function to set the bits
int setBits(int n, int m, int i, int j)
{
    // number with all 1's
    int allOnes = ~0;

    // Set all the bits in the left of j
    int left = allOnes << (j + 1);

    // Set all the bits in the right of j
    int right = ((1 << i) - 1);

    // Do Bitwsie OR to get all the bits 
    // set except in the range from i to j
    int mask = left | right;

    // clear bits j through i
    int masked_n = n & mask;

    // move m into the correct position
    int m_shifted = m << i;

    // return the Bitwise OR of masked_n 
    // and shifted_m
    return (masked_n | m_shifted);
}

// Drivers program
int main()
{
    int n = 2, m = 4;
    int i = 2, j = 4;
    cout << setBits(n, m, i, j);
    return 0;
}

Java

// Java Program 
public class GFG
{
    // Function to set the bits
    static int setBits(int n, int m, int  i, int j)
    {
        
        // number with all 1's
        int  allOnes = ~0;
        
        // Set all the bits in the left of j
        int left = allOnes << (j + 1);
        
        // Set all the bits in the right of j
        int right = ((1 << i) - 1);
        
        // Do Bitwise OR to get all the bits 
        // set except in the range from i to j
        int mask = left | right;
        
        // clear bits j through i
        int masked_n = n & mask;
        
        // move m into the correct position
        int m_shifted = m << i;
        
        // return the Bitwise OR of masked_n 
        // and shifted_m
        return (masked_n | m_shifted);
    }
    
    // Driver Program to test above function
    public static void main(String[] args) 
    {
        int n = 2, m = 4;
        int i = 2, j = 4;
        System.out.println(setBits(n, m, i, j));
    }
}
// This code is contributed by Sumit Ghosh

Python

# Python program for above implementation

# Function to set the bits
def setBits(n, m, i, j):

    # number with all 1's
    allOnes = not 0
 
    # Set all the bits in the left of j
    left = allOnes << (j + 1)
 
    # Set all the bits in the right of j
    right = ((1 << i) - 1)
 
    # Do Bitwsie OR to get all the bits 
    # set except in the range from i to j
    mask = left | right
 
    # clear bits j through i
    masked_n = n & mask
 
    # move m into the correct position
    m_shifted = m << i
 
    # return the Bitwise OR of masked_n 
    # and shifted_m
    return (masked_n | m_shifted)
 
# Drivers program
n, m = 2, 4
i, j = 2, 4
print setBits(n, m, i, j)

# This code is submitted by Sachin Bisht


Output:

18

Reference:
https://www.careercup.com/question?id=8863294

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